problem solving of quadratic inequalities

Solving Quadratic Inequalities

... and more ...

A Quadratic Equation (in Standard Form) looks like:

The above is an equation (=) but sometimes we need to solve inequalities like these:

Solving inequalities is very like solving equations ... we do most of the same things.

So this is what we do:

• find the "=0" points
• greater than zero (>0), or
• less than zero (<0)
• then pick a test value to find out which it is (>0 or <0)

Here is an example:

Example: x 2 − x − 6 < 0

x 2 − x − 6 has these simple factors (what luck!):

(x+2)(x−3) < 0

Firstly , let us find where it is equal to zero:

(x+2)(x−3) = 0

It is equal to zero when x = −2 or x = +3 because when x = −2, then (x+2) is zero or when x = +3, then (x−3) is zero

So between −2 and +3, the function will either be

• always greater than zero, or
• always less than zero

We don't know which ... yet!

Let's pick a value in-between (say x=0) and test it:

So between x=−2 and x=+3, the function is less than zero.

And that is the region we want, so...

x 2 − x − 6 < 0 in the interval (−2, 3)

Note: x 2 − x − 6 > 0   in the interval (−∞,−2) and (3, +∞)

Also try the Inequality Grapher .

What If It Doesn't Go Through Zero?

A "real world" example, a stuntman will jump off a 20 m building. a high-speed camera is ready to film him between 15 m and 10 m above the ground..

When should the camera film him?

We can use this formula for distance and time:

d = 20 − 5t 2

• d = distance above ground (m), and
• t = time from jump (seconds)

(Note: if you are curious about the formula, it is simplified from d = d 0 + v 0 t + ½a 0 t 2 , where d 0 =20 , v 0 =0 , and a 0 =−9.81 the acceleration due to gravity.)

OK, let's go.

First , let us sketch the question:

The distance we want is from 10 m to 15 m :

10 < d < 15

And we know the formula for d :

10 < 20 − 5t 2 < 15

Now let's solve it!

First, let's subtract 20 from both sides:

−10 < −5t 2 <−5

 Now multiply both sides by −(1/5). But because we are multiplying by a negative number, the inequalities will change direction ... read Solving Inequalities to see why.

2 > t 2 > 1

To be neat, the smaller number should be on the left, and the larger on the right. So let's swap them over (and make sure the inequalities still point correctly):

1 < t 2 < 2

 Lastly, we can safely take square roots, since all values are greater then zero:

√1 < t < √2

We can tell the film crew:

"Film from 1.0 to 1.4 seconds after jumping"

Higher Than Quadratic

The same ideas can help us solve more complicated inequalities:

Example: x 3 + 4 ≥ 3x 2 + x

First, let's put it in standard form:

x 3 − 3x 2 − x + 4 ≥ 0

This is a cubic equation (the highest exponent is a cube, i.e. x 3 ), and is hard to solve, so let us graph it instead:

The zero points are approximately :

And from the graph we can see the intervals where it is greater than (or equal to) zero:

• From −1.1 to 1.3, and
• From 2.9 on

In interval notation we can write:

Approximately: [−1.1, 1.3] U [2.9, +∞)

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How to Solve Quadratic Inequalities

Last Updated: June 4, 2020 References

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been viewed 182,639 times.

Factoring the Inequality

Determining the Roots of the Inequality

Plotting the Solution Set on a Number Line

Plotting the Solution Set on a Coordinate Plane

Community Q&A

You Might Also Like

• ↑ http://www.mathsisfun.com/algebra/quadratic-equation.html
• ↑ http://www.mathwarehouse.com/dictionary/B-words/what-is-a-binomial.php
• ↑ https://www.khanacademy.org/math/algebra-home/alg-quadratics/alg-quadratic-inequalities/v/quadratic-inequality-example-2
• ↑ http://www.purplemath.com/modules/ineqsolv.htm
• ↑ http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/inequalitiesrev4.shtml
• ↑ http://www.themathpage.com/aprecalc/roots-zeros-polynomial.htm
• ↑ http://www.virtualnerd.com/algebra-2/quadratics/inequalities/graphing-solving-inequalities/graph-inequality
• ↑ http://www.dummies.com/test-prep/act/act-trick-for-quadratics-how-to-quickly-find-the-direction-of-a-parabola/
• ↑ http://www.varsitytutors.com/hotmath/hotmath_help/topics/graphing-quadratic-inequalities
• ↑ https://www.khanacademy.org/math/algebra-home/alg-quadratics/alg-quadratic-inequalities/v/quadratic-inequalities-visual-explanation
• ↑ http://www.purplemath.com/modules/ineqquad.htm

About This Article

To solve a quadratic inequality, first write it as ax^2 + bx + c is less than 0. Then find 2 factors whose product is its first term and 2 factors whose product is its third term. Be sure the 2 factors whose product is its third term also have a sum that’s equal to its second term. Now determine whether your factors have the same or opposite signs by seeing if the product of the factors is greater or less than 0. Finally, turn each factor into an inequality, simplify, and check the validity of the roots for each option. If you want to learn how to show the solutions on a number line, keep reading the article! Did this summary help you? Yes No

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9.8 Solve Quadratic Inequalities

Learning objectives.

By the end of this section, you will be able to:

• Solve quadratic inequalities graphically
• Solve quadratic inequalities algebraically

Be Prepared 9.22

Before you get started, take this readiness quiz.

Solve: 2 x − 3 = 0 . 2 x − 3 = 0 . If you missed this problem, review Example 2.2 .

Be Prepared 9.23

Solve: 2 y 2 + y = 15 2 y 2 + y = 15 . If you missed this problem, review Example 6.45 .

Be Prepared 9.24

Solve 1 x 2 + 2 x − 8 > 0 1 x 2 + 2 x − 8 > 0 If you missed this problem, review Example 7.56 .

We have learned how to solve linear inequalities and rational inequalities previously. Some of the techniques we used to solve them were the same and some were different.

We will now learn to solve inequalities that have a quadratic expression. We will use some of the techniques from solving linear and rational inequalities as well as quadratic equations.

We will solve quadratic inequalities two ways—both graphically and algebraically.

Solve Quadratic Inequalities Graphically

A quadratic equation is in standard form when written as ax 2 + bx + c = 0. If we replace the equal sign with an inequality sign, we have a quadratic inequality in standard form.

Quadratic Inequality

A quadratic inequality is an inequality that contains a quadratic expression.

The standard form of a quadratic inequality is written:

The graph of a quadratic function f ( x ) = ax 2 + bx + c = 0 is a parabola. When we ask when is ax 2 + bx + c < 0, we are asking when is f( x ) < 0. We want to know when the parabola is below the x -axis.

When we ask when is ax 2 + bx + c > 0, we are asking when is f ( x ) > 0. We want to know when the parabola is above the x -axis.

Example 9.64

How to solve a quadratic inequality graphically.

Solve x 2 − 6 x + 8 < 0 x 2 − 6 x + 8 < 0 graphically. Write the solution in interval notation.

Try It 9.127

ⓐ Solve x 2 + 2 x − 8 < 0 x 2 + 2 x − 8 < 0 graphically and ⓑ write the solution in interval notation.

Try It 9.128

ⓐ Solve x 2 − 8 x + 12 ≥ 0 x 2 − 8 x + 12 ≥ 0 graphically and ⓑ write the solution in interval notation.

We list the steps to take to solve a quadratic inequality graphically.

Solve a quadratic inequality graphically.

• Step 1. Write the quadratic inequality in standard form.
• Step 2. Graph the function f ( x ) = a x 2 + b x + c . f ( x ) = a x 2 + b x + c .
• Step 3. Determine the solution from the graph.

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the x -axis but note how the position of the parabola affects the solution.

Example 9.65

Solve − x 2 − 8 x − 12 ≤ 0 − x 2 − 8 x − 12 ≤ 0 graphically. Write the solution in interval notation.

Try It 9.129

ⓐ Solve − x 2 − 6 x − 5 > 0 − x 2 − 6 x − 5 > 0 graphically and ⓑ write the solution in interval notation.

Try It 9.130

ⓐ Solve − x 2 + 10 x − 16 ≤ 0 − x 2 + 10 x − 16 ≤ 0 graphically and ⓑ write the solution in interval notation.

Solve Quadratic Inequalities Algebraically

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the critical points for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the critical points to divide the number line into intervals and then determine whether the quadratic expression willl be postive or negative in the interval. We then determine the solution for the inequality.

Example 9.66

How to solve quadratic inequalities algebraically.

Solve x 2 − x − 12 ≥ 0 x 2 − x − 12 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.131

Solve x 2 + 2 x − 8 ≥ 0 x 2 + 2 x − 8 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.132

Solve x 2 − 2 x − 15 ≤ 0 x 2 − 2 x − 15 ≤ 0 algebraically. Write the solution in interval notation.

In this example, since the expression x 2 − x − 12 x 2 − x − 12 factors nicely, we can also find the sign in each interval much like we did when we solved rational inequalities. We find the sign of each of the factors, and then the sign of the product. Our number line would like this:

The result is the same as we found using the other method.

We summarize the steps here.

Solve a quadratic inequality algebraically.

• Step 2. Determine the critical points—the solutions to the related quadratic equation.
• Step 3. Use the critical points to divide the number line into intervals.
• Step 4. Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
• Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Example 9.67

Solve - x 2 + 6 x − 7 ≥ 0 - x 2 + 6 x − 7 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.133

Solve − x 2 + 2 x + 1 ≥ 0 − x 2 + 2 x + 1 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.134

Solve − x 2 + 8 x − 14 < 0 − x 2 + 8 x − 14 < 0 algebraically. Write the solution in interval notation.

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation ax 2 + bx + c = 0. These two solutions then gave us either the two x- intercepts for the graph or the two critical points to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

For a quadratic equation of the form ax 2 + bx + c = 0, a ≠ 0 . a ≠ 0 .

The last row of the table shows us when the parabolas never intersect the x -axis. Using the Quadratic Formula to solve the quadratic equation, the radicand is a negative. We get two complex solutions.

In the next example, the quadratic inequality solutions will result from the solution of the quadratic equation being complex.

Example 9.68

Solve, writing any solution in interval notation:

ⓐ x 2 − 3 x + 4 > 0 x 2 − 3 x + 4 > 0 ⓑ x 2 − 3 x + 4 ≤ 0 x 2 − 3 x + 4 ≤ 0

We are to find the solution to x 2 − 3 x + 4 > 0 . x 2 − 3 x + 4 > 0 . Since for all values of x x the graph is above the x -axis, all values of x make the inequality true. In interval notation we write ( − ∞ , ∞ ) . ( − ∞ , ∞ ) .

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the x -axis—no part of it is below the x -axis.

We are to find the solution to x 2 − 3 x + 4 ≤ 0 . x 2 − 3 x + 4 ≤ 0 . Since for all values of x the graph is never below the x -axis, no values of x make the inequality true. There is no solution to the inequality.

Try It 9.135

Solve and write any solution in interval notation: ⓐ − x 2 + 2 x − 4 ≤ 0 − x 2 + 2 x − 4 ≤ 0 ⓑ − x 2 + 2 x − 4 ≥ 0 − x 2 + 2 x − 4 ≥ 0

Try It 9.136

Solve and write any solution in interval notation: ⓐ x 2 + 3 x + 3 < 0 x 2 + 3 x + 3 < 0 ⓑ x 2 + 3 x + 3 > 0 x 2 + 3 x + 3 > 0

Section 9.8 Exercises

Practice makes perfect.

In the following exercises, ⓐ solve graphically and ⓑ write the solution in interval notation.

x 2 + 6 x + 5 > 0 x 2 + 6 x + 5 > 0

x 2 + 4 x − 12 < 0 x 2 + 4 x − 12 < 0

x 2 + 4 x + 3 ≤ 0 x 2 + 4 x + 3 ≤ 0

x 2 − 6 x + 8 ≥ 0 x 2 − 6 x + 8 ≥ 0

− x 2 − 3 x + 18 ≤ 0 − x 2 − 3 x + 18 ≤ 0

− x 2 + 2 x + 24 < 0 − x 2 + 2 x + 24 < 0

− x 2 + x + 12 ≥ 0 − x 2 + x + 12 ≥ 0

− x 2 + 2 x + 15 > 0 − x 2 + 2 x + 15 > 0

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

x 2 + 3 x − 4 ≥ 0 x 2 + 3 x − 4 ≥ 0

x 2 + x − 6 ≤ 0 x 2 + x − 6 ≤ 0

x 2 − 7 x + 10 < 0 x 2 − 7 x + 10 < 0

x 2 − 4 x + 3 > 0 x 2 − 4 x + 3 > 0

x 2 + 8 x > − 15 x 2 + 8 x > − 15

x 2 + 8 x < − 12 x 2 + 8 x < − 12

x 2 − 4 x + 2 ≤ 0 x 2 − 4 x + 2 ≤ 0

− x 2 + 8 x − 11 < 0 − x 2 + 8 x − 11 < 0

x 2 − 10 x > − 19 x 2 − 10 x > − 19

x 2 + 6 x < − 3 x 2 + 6 x < − 3

−6 x 2 + 19 x − 10 ≥ 0 −6 x 2 + 19 x − 10 ≥ 0

−3 x 2 − 4 x + 4 ≤ 0 −3 x 2 − 4 x + 4 ≤ 0

−2 x 2 + 7 x + 4 ≥ 0 −2 x 2 + 7 x + 4 ≥ 0

2 x 2 + 5 x − 12 > 0 2 x 2 + 5 x − 12 > 0

x 2 + 3 x + 5 > 0 x 2 + 3 x + 5 > 0

x 2 − 3 x + 6 ≤ 0 x 2 − 3 x + 6 ≤ 0

− x 2 + x − 7 > 0 − x 2 + x − 7 > 0

− x 2 − 4 x − 5 < 0 − x 2 − 4 x − 5 < 0

−2 x 2 + 8 x − 10 < 0 −2 x 2 + 8 x − 10 < 0

− x 2 + 2 x − 7 ≥ 0 − x 2 + 2 x − 7 ≥ 0

Writing Exercises

Explain critical points and how they are used to solve quadratic inequalities algebraically.

Solve x 2 + 2 x ≥ 8 x 2 + 2 x ≥ 8 both graphically and algebraically. Which method do you prefer, and why?

Describe the steps needed to solve a quadratic inequality graphically.

Describe the steps needed to solve a quadratic inequality algebraically.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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• Quadratic Inequalities – Explanation & Examples

Solving Quadratic Inequalities – Explanation & Examples

A quadratic inequality is an equation of second degree that uses an inequality sign instead of an equal sign.

The solutions to quadratic inequality always give the two roots. The nature of the roots may differ and can be determined by discriminant (b 2 – 4ac).

The general forms of the quadratic inequalities are:

ax 2 + bx + c < 0

ax 2 + bx + c ≤ 0

ax 2 + bx + c > 0

ax 2 + bx + c ≥ 0

Examples of quadratic inequalities are:

x 2 – 6x – 16 ≤ 0, 2x 2  – 11x + 12 > 0, x 2  + 4 > 0, x 2  – 3x + 2 ≤ 0 etc.

How to Solve Quadratic Inequalities?

Examples of quadratic inequalities are: x 2 – 6x – 16 ≤ 0, 2x 2  – 11x + 12 > 0, x 2  + 4 > 0, x 2  – 3x + 2 ≤ 0 etc.

Solving a quadratic inequality in Algebra is similar to solving a quadratic equation. The only exception is that, with quadratic equations, you equate the expressions to zero, but with inequalities, you’re interested in knowing what’s on either side of the zero i.e. negatives and positives.

Quadratic equations can be solved by either the factorization method or by use of the quadratic formula . Before we can learn how to solve quadratic inequalities, let’s recall how quadratic equations are solved by handling a few examples.

How Quadratic Equations are Solved by Factorization Method?

Since we know we can similarly solve quadratic inequalities as quadratic equations, it is useful to understand how to factorize the given equation or inequality.

Let’s see a few examples here.

• 6x 2 – 7x + 2 = 0

⟹ 6x 2  – 4x – 3x + 2 = 0

Factorize the expression;

⟹ 2x (3x – 2) – 1(3x – 2) = 0

⟹ (3x – 2) (2x – 1) = 0

⟹ 3x – 2 = 0 or 2x – 1 = 0

⟹ 3x = 2 or 2x = 1

⟹ x = 2/3 or x = 1/2

Therefore, x = 2/3, ½

• Solve 3x 2 – 6x + 4x – 8 = 0

Factorize the expression on the left-hand side.

⟹ 3x 2  – 6x + 4x – 8 = 0

⟹ 3x (x – 2) + 4(x – 2) = 0

⟹ (x – 2) (3x + 4) = 0

⟹ x – 2 = 0 or 3x + 4 = 0

⟹ x = 2 or x = -4/3

Therefore, the roots of the quadratic equation are, x = 2, -4/3.

• Solve 2(x 2 + 1) = 5x

2x 2 + 2 = 5x

⟹ 2x 2  – 5x + 2 = 0

⟹ 2x 2  – 4x – x + 2 = 0

⟹ 2x (x – 2) – 1(x – 2) = 0

⟹ (x – 2) (2x – 1) = 0

⟹ x – 2 = 0 or 2x – 1 = 0

⟹ x = 2 or x = 1/2

Therefore, the solutions are x = 2, 1/2.

• (2x – 3) 2 = 25

Expand and factorize the expression.

(2x – 3) 2  = 25

⟹ 4x 2  – 12x + 9 – 25 = 0

⟹ 4x 2  – 12x – 16 = 0

⟹ x 2  – 3x – 4 = 0

⟹ (x – 4) (x + 1) = 0

⟹ x = 4 or x = -1

• Solve x 2 + (4 – 3y) x – 12y = 0

Expand the equation;

x 2  + 4x – 3xy – 12y = 0

⟹ x (x + 4) – 3y (x + 4) = 0

x + 4) (x – 3y) = 0

⟹ x + 4 = 0 or x – 3y = 0

⟹ x = -4 or x = 3y

Thus, x = -4 or x = 3y

To solve a quadratic inequality, we also apply the same method as illustrated in the procedure below:

• Write the quadratic inequality in standard form: ax 2  + bx + c where a, b and are coefficients and a ≠ 0
• Determine the roots of the inequality.
• Write the solution in inequality notation or interval notation.

Solve the inequality x 2  – 4x > –3

First, make one side one side of the inequality zero by adding both sides by 3.

x 2  – 4x > –3 ⟹ x 2  – 4x + 3 > 0

Factor the left side of the inequality.

x 2  – 4x + 3 > 0 ⟹ (x – 3) (x – 1) > 0

Solve for all the zeroes for the inequality;

For, (x – 1) > 0 ⟹ x > 1 and for, (x – 3) > 0 ⟹ x>3

Since y is positive, we therefore choose the values of x which the curve will be above the x-axis. x < 1 or x > 3

Solve the inequality x 2  – x > 12.

To write the inequality in standard form, subtract both sides of the inequality by 12.

x 2  – x > 12 ⟹ x 2  – x – 12 > 0.

Factorize the quadratic inequality to get to;

( x  – 4) ( x  + 3) > 0

For, (x + 3) > 0 ⟹ x > -3

For x – 4 > 0 ⟹ x > 4

The values x < –3 or x > 4 are therefore the solution of this quadratic inequality.

Solve 2x 2  < 9x + 5

Write the inequality in standard form by making one side of the inequality zero.

2x 2  < 9x + 5 ⟹ 2x 2  – 9x – 5 < 0

Factor the left side of the quadratic inequality.

2x 2  – 9x – 5 < 0 ⟹ (2x + 1) (x – 5) < 0

Solve for all the zeroes for the inequality

For, (x – 5) < 0 ⟹ x < 5 and for (2x + 1) < 0 ⟹ x < -1/2

Since y is negative for the equation 2x 2  – 9x – 5 < 0, we therefore choose the values of x which the curve will be below the x axis.

Therefore, the solution is   -1/2 < x < 5

Solve – x 2 + 4 < 0.

Since the inequality is already in standard form, we therefore factor the expression.

-x 2 + 4 < 0 ⟹ (x + 2) (x – 2) < 0

For, (x + 2) < 0 ⟹ x < -2 and for, (x – 2) < 0 ⟹ x < 2

The y for –x 2 + 4 < 0 is negative; therefore, we choose the values of x in which the curve will below the x- axis: –2 < x > 2

Solve 2x 2  + x − 15 ≤ 0.

Factor the quadratic equation.

2x 2  + x − 15 = 0

2x 2  + 6x – 5x− 15 = 0

2x (x + 3) – 5(x + 3) = 0

(2x – 5) (x + 3) = 0

For, 2x – 5 = 0 ⟹ x= 5/2 and for, x + 3= 0 ⟹ x = -3

Since the y for 2x 2  + x − 15 ≤ 0 is negative, the we choose the values of x in which the curve will be below the x axis. Therefore, x ≤ -3 or x ≥5/2 is the solution.

Solve – x 2  + 3x − 2 ≥ 0

Multiply the quadratic equation by -1 and remember to change the sign.

x 2  – 3x + 2 = 0

x 2  – 1x – 2x + 2 = 0

x (x – 1) – 2(x – 1) = 0

(x – 2) (x – 1) = 0

For, x – 2 = 0 ⟹ x = 2 and for, x – 1= 0 ⟹x=1

Therefore, the solution to the quadratic inequality is 1 ≤ x ≤ 2

Solve x 2 − 3x + 2 > 0

Factorize the expression to get;

x 2 − 3x + 2 > 0 ⟹ (x − 2) (x − 1) > 0

Now solve for the roots of the inequality as;

(x − 2) > 0 ⟹ x > 2

(x − 1) > 0 ⟹x > 1

The curve for x 2 − 3x + 2 > 0 has positive y, therefore which choose the values of x in which the curve will be above the x-axis. The solution is hence, x < 1 or x > 2.

Solve −2x 2 + 5x + 12 ≥ 0

Multiply the entire expression by -1 and change the inequality sign

−2x 2 + 5x + 12 ≥ 0 ⟹2x 2 − 5x − 12 ≤ 0

(2x + 3) (x − 4) ≤ 0.

Solve the roots;

(2x + 3) ≤ 0 ⟹ x ≤ -3/2.

(x − 4) ≤ 0 ⟹ x ≤ 4.

By applying the rule; (x – a) (x – b) ≥ 0, then a ≤ x ≤ b, we can comfortably write the solutions of this quadratic inequality as:

-3/2 ≤ x ≤ 4.

x 2  − x − 6 < 0

Factorize x 2  − x − 6 to get;

(x + 2) (x − 3) < 0

Find the roots of the equation as;

(x + 2) (x − 3) = 0

x = −2 or x = +3 Because y is negative for x 2  − x − 6 < 0, then we choose an interval in which the curve will be below the x axis. Therefore, -2 < x < 3 is the solution.

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Quadratic inequalities

Quadratic Inequalities

Here we will learn about quadratic inequalities including how to solve quadratic inequalities, identify solution sets using inequality notation and represent solutions on a number line.

There are also quadratic inequalities worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What are quadratic inequalities?

Quadratic inequalities are similar to quadratic equations and when plotted they display a parabola. We can solve quadratic inequalities to give a range of solutions.

For example,

The quadratic equation x^{2}+ 6x +5 = 0 has two solutions .

This is shown on the graph below where the parabola crosses the x axis.  

We could solve this by factorising: (x + 1)(x + 5) = 0 .

The two values of x that equate this equation to zero are x = - 1 and x = - 5 .

The quadratic inequality x^{2}+ 6x +5  \leq 0 can be solved by factorising but instead of two solutions, there are a range of solutions .

x^{2}+ 6x +5 \leq 0 means that the we need to know the x values that when the graph is less than 0 .

This corresponds to where the curve is below the x axis .

We can see the from the graph that the x values need to be between -1 and -5 .

Therefore the solution to this inequality can be written using inequality notation,

-5\leq x \leq-1 .

The quadratic inequality x^{2} + 6x +5 > 0 is similar to the previous inequality and has a range of solutions.

x^{2}+ 6x +5 > 0 means that we need to know the x values when the graph is greater than 0 . 

This corresponds to where the curve is above the x axis .

We can see the from the graph that the x values need to be greater than -1 and less than -5 .

Therefore the solution to this inequality can be written using inequality notation as two inequalities  x > -1 or x < -5 .

How to solve quadratic inequalities

In order to solve quadratic inequalities by factorising:

Factorise the quadratic expression.

Find the values of x that make each bracket equal zero.

Write the solution using inequality notation.

In order to solve quadratic inequalities by using the quadratic formula:

Identify values of a, b and c to substitute into the quadratic formula.

• Solve the quadratic inequality using the quadratic formula.

Simplify to calculate the solutions of the inequality.

How to solve quadratic inequalities by factorising

How to solve quadratic inequalities by using the quadratic formula

Quadratic inequalities worksheet

Get your free quadratic inequalities worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on inequalities

Quadratic inequalities is part of our series of lessons to support revision on inequalities . You may find it helpful to start with the main inequalities lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Inequalities
• Solving inequalities  
• Inequalities on a number line
• Inequalities on a graph
• Linear inequalities
• Greater than sign
• Less than sign

Quadratic inequalities examples

Example 1: solving by factorising.

Solve x ^{2}+7x+10 < 0 .

2 Find the values of x that make each bracket equal zero.

(x + 2) = 0 when x = -2

(x + 5) = 0 when x = -5

3 Write the solution using inequality notation.

We need the values of x that produce a graph that is less than 0 and so below the x axis. 

The inequality is given by -5 < x < -2 .

4 Represent the solution set on a number line.

-5 and -2 are not included in the solution set so these values are indicated with open circles.

Example 2: solving by factorising

Solve x^{2}+3x+10\leq0 .

(x - 2) = 0 when x = 2

We need the values of x that produce a graph that is less than or equal to 0 and so below the x axis.

The inequality is given by -5\leq x \leq2 .

Represent the solution set on a number line.

-5 and 2 are included in the solution set so these values are indicated with closed circles.

Example 3: solving by factorising

Solve x^{2}-6x+8\geq0 .

Find the values of x that make each bracket equal zero .

(x -2) = 0 when x = 2

(x -4 ) = 0 when x = 4

We need the values of x that produce a graph that is greater than or equal to 0 and so above the x axis.

There will be two separate inequalities as there are two separate regions indicated on the graph.

x \geq4 and x \leq2 .

2 and 4 are included in the solution set so these values are indicated with closed circles.

Example 4: solving by factorising

List the integers that satisfy 2x^{2} + 9x + 4\leq0

(2x + 1) = 0 when x = -\frac{1}{2}

(x +4 ) = 0 when x = -4

We need the values of x that produce a graph that is less than or equal to 0 and so below the x axis. 

The inequality is given by -4\leq x \leq-\frac{1}{2} .

List the integer values that are included in the solution set.

x is any value greater than or equal to  -4 and less than or equal to -\frac{1}{2} . The highest integer below -\frac{1}{2} is -1 .

Example 5: solving by factorising

Solve x^{2}-9>0 .

(x +3) = 0 when x = -3

(x -3 ) = 0 when x = 3

We need the values of x that produce a graph that is greater than 0 and so above the x axis.

x>3 and x < -3 .

-3 and 3 are not included in the solution set so these values are indicated with closed circles.

Example 6: solving by factorising and rearranging

List the integers that satisfy x^{2}<16 .

Rearrange the inequality.

Rearrange the inequality so that it is less than 0 by subtracting 16 from both sides.

Factorise the inequality.

(x +4) = 0 when x = -4

Write the solution using inequality notation .

The inequality is given by -4 < x < 4 .

x is any value greater than -4 and less than 4 .

Example 7: solving by using the quadratic formula

Solve 5x^{2}+6x-12\leq0 .

A quadratic expression is in the form ax^{2} + bx  + c .

In this inequality a = 5, b = 6 and c = -12 .

Substitute the values into the quadratic formula.

The quadratic formula is x = \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} .

Substituting the values a = 5, b = 6 and c = -12 gives x = \frac{-6 \pm \sqrt{6^{2}-4 \times 5 \times-12}}{2 \times 5} .

We need to use the quadratic formula to solve rather than factorising as the discriminant is not a square number.

In this case the answer is given in surd form but you may be asked to leave the answers rounded to a degree of accuracy.

We need the values of x that produce a graph that is less than or equal 0 and so below the x axis.

The inequality is given by \frac{-3- \sqrt{69}}{5} \leq x \leq \frac{-3+ \sqrt{69}}{5} .

Example 8: solving by using the quadratic formula

Solve 2x^{2}-3x- 4 < 0 .

A quadratic expression is in the form ax^{2}+bx+c .

In this inequality a = 2, b = -3 and c = -4 .

Substituting the values   a = 2, b =-36 and c = -4 gives

x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \times 2 \times-4}}{2 \times 2} .

In this case the answer is being given to two decimal places but you could leave the solution in surd form.

We need the values of x that produce a graph that is less than 0 and so below the x axis.

The inequality is given by -0.85 < x< 2.35 .

Rounded to two decimal places. 

Example 9: one solution

Solve x^{2}+6x+9\leq0 .

(x + 3)(x + 3) \leq 0 or (x+3)^{2}

As both brackets are equal, there is only one solution.

We need the values of x that produce a graph that is less than or equal to 0 and so below the x axis. This only happens when, x = -3 .

Common misconceptions

• Square root of a number has two solutions

A common error when calculating the square root of a number is only writing the positive solution. A negative number squared will also give a positive solution.

When solving x^2=16, x=\pm \sqrt{16} =\pm 4 .

• Not maintaining the original inequality sign

A common error is to accidentally change the direction of the inequality sign or not writing an inequality sign in the solution.

• Writing a double inequality when the solution set is split

When solving  x^{2} + 10x +16  < 0 there is one solution set which is below the x axis. Therefore the solution is -8 < x < -2 .

When solving  x^{2} + 10x +16  > 0 there is one solution set which is above the x axis. Therefore the solution is x<-8 and x > -2 .

Practice quadratic inequalities questions

1. Solve x^{2}+13x+30<0 .

Factorise the quadratic, (x + 3)(x +10) < 0 .

Find the values of x that make the bracket equal zero, x = -10 and x = -3 .

Write the solution using interval notation, -10 < x < -3 .

2. Solve x^{2}+3x-10\leq0 .

Factorise the quadratic, (x + 5)(x -2) \leq 0 .

Find the values of x that make the bracket equal zero, x = -5 and x = 2 .

Write the solution using interval notation, -5\leq x \leq2 .

3. Solve x^{2}-7x+6\geq0 .

x \leq 1 and x\geq 6

Factorise the quadratic, (x-6)(x-1)\geq 0 .

Find the values of x that make the bracket equal zero, x = 6 and x =1 .

Write the solution using interval notation, x \leq 1 and x\geq6 .

4. Solve 2x^{2}-11x+ 5 \leq 0 .

Factorise the quadratic, (2x-1)(x-5) \leq 0 .

Find the values of x that make the bracket equal zero, x =\frac{1}{2} and x =5 .

Write the solution using interval notation, \frac{1}{2} \leq x\leq 5 .

5. Solve x^{2}-25>0 .

x = -5 and x =5

Factorise the quadratic, (x + 5)(x -5) > 0 .

Find the values of x that make the bracket equal zero, x = -5 and x =5 .

Write the solution using interval notation, x < -5 and x > 5 .

6. Solve   x^{2}<1 .

Rearrange to make the inequality less than zero, x^{2} – 1<0 .

Factorise the quadratic, (x + 1)(x -5) < 0 .

Find the values of x that make the bracket equal zero, x = -1 and x =1 .

Write the solution using interval notation, -1 < x < 1 .

7. Solve 3x^{2}+4x-2\leq 0 .

3x^{2} + 4x – 2 \leq 0 .

Identify a, b and c from the form ax^{2} +bx + c \leq 0 , a = 3, b = 4, c = -2 .

Substitute the values in to the quadratic formula, x = \frac{-4 \pm \sqrt{4^{2}-4 \times 3 \times-2}}{2 \times 3} .

Simplify the inequality, x = \frac{-4 \pm \sqrt{16–24}}{6} .

Simplify further, x=\frac{-4 \pm \sqrt{40}}{6} .

You could simplify to surd form, x = \frac{-4 \pm 2 \sqrt{10}}{6} .

Simplify the fraction. You could leave in surd form, x = \frac{-2 \pm \sqrt{10}}{3} .

Correct to 2 decimal places, x \leq 0.39 and x \geq -1.72 .

Written in surd form written in interval notation, \frac{-2+\sqrt{10}}{3} \leq x \leq \frac{-2+\sqrt{10}}{3} .

Written as an inequality with values rounded to 2 decimal places, -1.72 \leq x \leq 0.39 .

8. Solve x^{2}-10x+25 \leq 0 .

x^{2} -10x +25 \leq 0 .

Factorise the quadratic, (x – 5)(x -5) \leq 0 .

Find the values of x that make the bracket equal zero, x =5 .

The quadratic only has one root at 5, so is never less than zero but equal to zero at 5, x = 5 .

Quadratic inequalities GCSE questions

1. Solve the inequality x^{2}-8x+15\leq 0 .

Factorise the inequality. 1 mark if there are one or two errors.

Roots found x = 3 and x = 5

Correct answer.

2. (a) Find the interval for which x^{2}+3x – 10 < 0 .

(b) Represent your solution to part (a) on the number line below.

1 mark if there are one or two errors.

Roots found x =2 and x = -5

-5 and 2 indicated.

Correct solution with open circles.

3. Here is a rectangle.

All measurements are in centimetres.

x is an integer.

The total area of the rectangle is less than 21cm^{2} .

Show that 4 <x < 7 .

x^{2} – 4x <21 .

x^{2} – 4x – 21< 0 .

(x – 7)(x + 3) < 0 .

4<x < 7 as you cannot have a negative value for distance.

Learning checklist

You have now learned how to:

• Solve quadratic inequalities
• Use inequality notation and represent solutions on a number line
• List integers that are in the solution set

The next lessons are

• Functions in algebra
• Laws of indices

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Quadratic Inequalities

Quadratic inequalities can be derived from quadratic equations. The word “quadratic” comes from the word “quadrature”, which means "square" in Latin. From this, we can define quadratic inequalities as second-degree inequation. Here, we first define a quadratic equation. The general form of a quadratic equation is ax 2 + bx + c = 0. Further if the quadratic polynomial ax 2 + bx + c is not equal to zero, then they are either ax 2 + bx + c > 0, or ax 2 + bx + c < 0, and are referred as quadratic inequalities.

In this mini-lesson, we will learn about quadratic inequality, solving quadratic inequalities, quadratic inequalities in one variable, quadratic inequalities formula, and the graph of quadratic inequality, with the help of examples, FAQs.

What Do You Mean By Quadratic Inequalities?

The quadratic inequality is a second-degree expression in x and has a greater than (>) or lesser than (<) inequality. the quadratic inequality has been derived from the quadratic equation ax 2 + bx + c = 0. Let us check the definition of quadratic inequality, the standard form, and the examples of quadratic inequalities.

If a quadratic polynomial in one variable is less than or greater than some number or any other polynomial (with a degree less than or equal to 2), then it is said to be a quadratic inequality. The difference between a quadratic equation and a quadratic inequality is that the quadratic equation is equal to some number while quadratic inequality is either less than or greater than some number. Some examples of quadratic inequalities in one variable are:

• x 2 + x - 1 > 0
• 2x 2 - 5x - 2 > 0
• x 2 + 2x - 1 < 0

Standard Form

The standard form of quadratic inequalities in one variable is almost the same as the standard form of a quadratic equation. The only difference is that the quadratic equation has an "equal to" sign in it while a quadratic inequality has a "greater than" or "less than" sign (> or <). The standard form of quadratic inequality can be represented as:

Example of Quadratic Inequality

Now, consider the scenario where you want to build a rectangular house with a length equal to two units more than twice its breadth. If you don't want the floor area of the house to be more than 1500 ft 2 , what length and breadth can you consider?

You know that the area of a rectangle is length times its breadth. Hence, the area of the house is (2 + 2x)x = 2x 2 + 2x, where x is the breadth of the rectangular house. Now, we know that the area cannot exceed 1500 ft 2 . Thus, the quadratic inequality for the above scenario is as follows.

2x 2 + 2x < 1500

How to solve quadratic inequalities.

Solving a quadratic inequality means to find the values of x which satisfy the given condition of the question. A quadratic second degree equation ax 2 + bx + c = 0 can have maximum 2 values of x. But a quadratic inequality can have more than 2 values. It can have infinite values of x which satisfy the condition ax 2 + bx + c < 0 or ax 2 + bx + c > 0. Solving a quadratic inequation means finding the range of values of x.

Now consider a quadratic expression ax 2 + bx + c. We can write the quadratic expression in the form of (x - α)(x - β) and α < β. Using the number line method the solution for the quadratic inequality can be expressed as follows.

It means that if ax 2 + bx + c > 0, then x can take values between - ∞ to α and β to +∞.

If x 2 + bx + c > 0, then x ∈ (-∞ , α) U (β, + ∞)

If x 2 + bx + c < 0, then x can take values between α and β.

If x 2 + bx + c < 0, then x ∈ (α, β)

Let's take a quadratic inequality x 2 - 1 > 0. Here the expression x 2 - 1 > 0 can be factorized as (x - 1)(x + 1) > 0. This gives the values of α = -1 and β = 1. Hence, we obtain the range of x as x ∈ (-∞ , -1) U (1, + ∞)

If the quadratic inequality is x 2 - 1 < 0. The expression x 2 - 1 < 0, can be factorized as (x - 1)(x + 1) < 0. This gives α = -1 and β = 1. Therefore, the range of x is x ∈ (-1, 1)

If the quadratic inequality is x 2 - 1 > 0 (where it shows the quadratic inequality is greater than or equal to zero). The expression x 2 - 1 > 0 can be factorized as (x - 1)(x + 1) > 0. Here we obtain α = -1 and β = 1, and the range of x is x ∈ {-∞ , -1] U [+1, + ∞}

If the quadratic inequality is x 2 - 1 < 0 (where it shows the quadratic inequality is less than or equal to zero). The expression x 2 - 1 < 0 is factorized as (x - 1)(x + 1) < 0. Here the roots of the expression are α = -1 and β = 1, and the range of x is x ∈ [-1, +1]

Notations Used In Quadratic Inequalities

The notation of greater than (>) and lesser. than (< ) is often used in quadratic equations. The quadratic equation ax 2 + bx + c = 0 is written as a quadratic equation by replacing equals to the symbol (=) with greater than or lesser than inequality. The general format of a quadratic inequality is ax 2 + bx + c > 0, or ax 2 + bx + c < 0. Further, let us check some of the other important notations used in quadratic inequalities.

• ( ) → Open Brackets
• [ ] → Closed Brackets
• o → Open Value( x cannot take this value)
• • → Closed Value( x can take this value)
• (-1, 1) → x cannot take value -1 and 1.
• [-1, 1) → x can take value -1 but not 1.
• (-1, 1] → x cannot take value -1 but it can take value 1.
• [-1, 1] → x can take both -1 and 1 values

☛ Related Topics

• Cube Root Function
• Piecewise Function
• Nonlinear Function
• Probability Mass Function
• Objective Function

Solved Examples on Quadratic Inequalities

Example 1: Find the range of values of x which satisfy the quadratic inequality x 2 - 7x + 10 < 0.

First let's factorize the quadratic expression x 2 - 7x + 10.

x 2 - 7x + 10 < 0 x 2 - 5x - 2x + 10 < 0 x(x - 5) - 2(x - 5) < 0 (x - 2)(x - 5) < 0

Hence, the values of x that satisfy the quadratic inequality are x ∈ (2, 5)

Therefore we have x ∈ (2, 5).

Example 2: Help Mathew find the range of values of m such that the quadratic inequality x 2 - mx + m > 0 is true for all values x.

For the quadratic expression x 2 - mx + m, the coefficient of x 2 is positive, and this expression is always positive and equal to zero. Hence, the discriminant of the quadratic equation is always negative and equal to zero. The discriminant of the equation is (-m) 2 - 4(1)(m)

Hence we have the following inequality.

(-m) 2 - 4m < 0

(-m) 2 - 4m < 0 m^2 - 4m < 0 m(m - 4) < 0

Thus, the values of m that satisfy the conditions for quadratic inequality are m ∈ [0, 4]

Therefore the solution of the quadratic inequality is m ∈ [0, 4].

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Practice Questions on Quadratic Inequalities

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FAQs on Quadratic Inequalities

How do you explain solving quadratic inequalities.

Quadratic inequalities can be solved by first factorizing the quadratic expression, then finding the range of values of x which satisfy the quadratic inequality. You can follow this section for more details on solving quadratic inequalities.

How do you solve quadratic inequalities in two variables?

An quadratic inequalities in two variables can be solved by drawing the quadratic inequalities graph of the expression. For example, the solutions of the inequality y - x 2 + 1 > 0 can be solved by drawing the quadratic inequalities graph of y = x 2 - 1.

From the graph, we can clearly observe that the range of x is x ∈ (-∞, +∞) and the range of y is y ∈ (-1, +∞)

How do you solve quadratic inequalities equal to zero?

If a quadratic inequality, let's say x 2 -5x - 6 > 0 is solved by equating to zero, it means that now we have to solve a quadratic equation x 2 -5x - 6 = 0. The quadratic equation x 2 -5x + 6 = 0 can have maximum two roots. The two values of x which satisfy the above quadratic equations are x = 2 and x = 3.

What conditions don't affect the direction of Quadratic inequality?

Adding and subtracting any number on both sides of the inequality doesn't affect the direction of the inequality. Multiplying or dividing both sides of the inequality by a negative number changes the direction of the inequality. Multiplying or dividing both sides of the inequality by a positive number doesn't affect the direction of the inequality.

How are quadratic equations solved by the factorization method?

To solve a quadratic equation with the help of the factorization method, we split the middle term (the term with x) of the equation such that when multiplying the two numbers obtained, we get the constant term. For example, the roots of the quadratic equation x 2 -3x + 2 = 0 by factorization method can be obtained in the following way:

x 2 - 3x + 2 = 0 x 2 - 2x - x + 2 = 0 x(x - 2) -1(x - 2) = 0 (x - 1)(x - 2) = 0

Hence, x = 1 and x = 2 are the two roots of the quadratic equation.

What is the first step in solving quadratic inequality?

The first step while solving a quadratic inequality is to factorize the quadratic expression.

How do you solve quadratic inequalities word problems?

To solve a quadratic inequality word problem, first, deduce the quadratic inequality from the word problem, then solve it according to the conditions given in the question.

Can you square a quadratic inequality?

Usually, we do not prefer squaring the quadratic inequality because it might change the inequality, but for the case in which both the sides are always positive, then we can square both sides. For example, in the quadratic inequality x 2 > 0, both sides of the inequality are always positive. Hence, we can square it. x 4 > 0 this does not change the inequality.

How do you know if the given inequality is a quadratic inequality or not?

If the expression of the inequality is a quadratic polynomial, it will be a quadratic inequality.

What do you mean by the quadratic inequalities formula?

The quadratic inequalities formula for a quadratic equation is \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). We can factorize the quadratic expression with the help of the above quadratic formula. Then we can solve the inequality.

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Last modified on May 13th, 2024

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Quadratic inequalities.

If we replace a quadratic equation’s equality sign (=) in the standard form ax 2 + bx + c = 0 with an inequality sign, it becomes a quadratic inequality.

Here are a few examples of quadratic inequalities:

5x 2 – 11x + 6 > 0

x 2 + 5x – 6 < 0

Standard Forms

Depending on the sign, the 4 standard forms of quadratic inequality are:

• ax 2 + bx + c > 0
• ax 2 + bx + c < 0
• ax 2 + bx + c ≥ 0
• ax 2 + bx + c ≤ 0

Here, like a quadratic equation, ‘a’ (≠ 0), ‘b,’ and ‘c’ are the constants, and ‘x’ is a variable.

Solving quadratic inequalities involves finding the value of the variable that satisfies the inequality. It can be done both graphically and algebraically.

Graphically

While solving quadratic inequalities graphically, we find the intervals where the inequality is either above or below the x-axis, depending on the inequality sign.

To understand the concept, let us solve the inequality 5x 2 – 11x + 6 > 0.

Step 1. Writing in Standard Form

5x 2 – 11x + 6 > 0, which is in its standard form (ax 2 + bx + c > 0).

Step 2. Graphing the Quadratic Function

Finding the Line of Symmetry

${x=-\dfrac{b}{2a}}$

⇒ ${x=-\dfrac{-11}{2\times 5}}$

⇒ ${x=\dfrac{11}{10}}$

Finding the Vertex of the Quadratic Function 

f(x) = 5x 2 – 11x + 6, 

${f\left( \dfrac{11}{10}\right) =5\left( \dfrac{11}{10}\right) ^{2}-11\left( \dfrac{11}{10}\right) +6}$

⇒ ${f\left( \dfrac{11}{10}\right) =\dfrac{605}{100}-\dfrac{121}{10}+6}$

⇒ ${f\left( \dfrac{11}{10}\right) =\dfrac{-5}{100}}$

⇒ ${f\left( \dfrac{11}{10}\right) =\dfrac{-1}{20}}$

Thus, the vertex is ${\left( \dfrac{11}{10},\dfrac{-1}{20}\right)}$ ⇒ (1.1, -0.05)

Finding the x-intercepts

By putting f(x) = 0, we get

5x 2 – 11x + 6 = 0

⇒ 5x 2 – 5x – 6x + 6 = 0

⇒ (x – 1)(5x – 6) = 0

⇒ x = 1, ${\dfrac{6}{5}}$

⇒ x = 1, 1.2

Thus, x-intercepts are (1, 0) and (1.2, 0)

Now, graphing the quadratic function f(x) = 5x 2 – 11x + 6, we get

Step 4. Determining the Solution Set 

Since the inequality has a ‘>’ sign, the vertices are not included, and the solution set is (-∞, 1) ∪ (1.2, ∞). On graphing the solution, we get

Algebraically

Let us solve the inequality x 2 + 5x – 6 < 0

x 2 + 5x – 6 < 0, which is in its standard form of ax 2 + bx + c < 0.

Step 2. Determining the Critical Points

Factoring the inequality, we get

⇒ x 2 + 6x – x – 6 < 0

⇒ (x + 6)(x – 1) < 0

Now, finding the critical points (the solutions) to the related quadratic equation, we get

(x + 6)(x – 1) = 0

⇒ x = -6, 1

Step 3. Dividing the Number Line into Intervals with the Critical Points

Step 4. Determining the Sign of the Expression for Distinct Intervals

At x = -10 (< -6), (-10) 2 + 5(-10) – 6 = 44 ≮ 0

At x = -1 (> -6 but < 1), (-1) 2 + 5(-1) – 6 = -10 < 0

At x = 5 (> 1), (5) 2 + 5(5) – 6 = 44 ≮ 0

Thus, at -6 < x < 1, the inequality holds.

Step 5. Representing the Solution Set on the Number Line

Here, the solution set of the quadratic inequality is (-6, 1), the interval notation.

Solved Examples

The length of a rectangular box is 18 meters more than thrice its width. Find all possible rectangle widths that would result in an area of no more than 120 square meters and graph the solution on a number line.

Let ‘l’ and ‘w’ be the length and width of the rectangular box. As we know, the length is 18 meters, more than thrice the width Here, l = 18 + 3w Area = l ⋅ w =  w(18 + 3w) = 18w + 3w 2 Since the area cannot exceed 120 m 2 Thus, 18w + 3w 2 ≤ 120 ⇒ 3w 2 + 18w – 120 ≤ 0 ⇒ w 2 + 6w – 40 ≤ 0 ⇒ w 2 + 10w – 4w – 40 ≤ 0 ⇒ (w + 10)(w – 4) ≤ 0 Now, finding the critical points (the solutions) to the related quadratic equation, we get (w + 10)(w – 4) = 0 ⇒ w = -10, 4 Here, the interval points are -10 and 4. Now, at w = -11 (< -10), 3(-11) 2 + 18(-11) = 363 – 198 = 165, which is not less than 120 At w = 0 (> -10 but < 4), 3(0) 2 + 18(0) = 0, which is less than 120 At w = 5 (> 4), 3(5) 2 + 18(5) = 165, which is not less than 120 Thus, at -10 < w < 4, the inequality holds. Since the width cannot be negative, the only valid solution is 0 < w ≤ 4 Thus, all possible rectangle widths resulting in an area of no more than 120 square meters are w ≤ 4 meters, which means w = {1, 2, 3, 4}

As we know, the y-intercept is 2, not -2 The options c) and d) are eliminated. Now, on factoring the quadratic expressions,  x 2 + 3x + 2 = x 2 + 2x + x + 2 = (x + 2)(x + 1) The x-intercepts are (-1, 0) and (-2, 0) Since the shaded part is above the curve. Thus, a) is the correct quadratic inequality, which represents the given graph.

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9.9: Solve Quadratic Inequalities

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Learning Objectives

By the end of this section, you will be able to:

• Solve quadratic inequalities graphically
• Solve quadratic inequalities algebraically

Be Prepared 9.22

Before you get started, take this readiness quiz.

Solve: 2 x − 3 = 0 . 2 x − 3 = 0 . If you missed this problem, review Example 2.2.

Be Prepared 9.23

Solve: 2 y 2 + y = 15 2 y 2 + y = 15 . If you missed this problem, review Example 6.45.

Be Prepared 9.24

Solve 1 x 2 + 2 x − 8 > 0 1 x 2 + 2 x − 8 > 0 If you missed this problem, review Example 7.56.

We have learned how to solve linear inequalities and rational inequalities previously. Some of the techniques we used to solve them were the same and some were different.

We will now learn to solve inequalities that have a quadratic expression. We will use some of the techniques from solving linear and rational inequalities as well as quadratic equations.

We will solve quadratic inequalities two ways—both graphically and algebraically.

Solve Quadratic Inequalities Graphically

A quadratic equation is in standard form when written as ax 2 + bx + c = 0. If we replace the equal sign with an inequality sign, we have a quadratic inequality in standard form.

Quadratic Inequality

A quadratic inequality is an inequality that contains a quadratic expression.

The standard form of a quadratic inequality is written:

a x 2 + b x + c < 0 a x 2 + b x + c ≤ 0 a x 2 + b x + c > 0 a x 2 + b x + c ≥ 0 a x 2 + b x + c < 0 a x 2 + b x + c ≤ 0 a x 2 + b x + c > 0 a x 2 + b x + c ≥ 0

The graph of a quadratic function f ( x ) = ax 2 + bx + c = 0 is a parabola. When we ask when is ax 2 + bx + c < 0, we are asking when is f( x ) < 0. We want to know when the parabola is below the x -axis.

When we ask when is ax 2 + bx + c > 0, we are asking when is f ( x ) > 0. We want to know when the parabola is above the x -axis.

Example 9.64

How to solve a quadratic inequality graphically.

Solve x 2 − 6 x + 8 < 0 x 2 − 6 x + 8 < 0 graphically. Write the solution in interval notation.

Try It 9.127

ⓐ Solve x 2 + 2 x − 8 < 0 x 2 + 2 x − 8 < 0 graphically and ⓑ write the solution in interval notation.

Try It 9.128

ⓐ Solve x 2 − 8 x + 12 ≥ 0 x 2 − 8 x + 12 ≥ 0 graphically and ⓑ write the solution in interval notation.

We list the steps to take to solve a quadratic inequality graphically.

Solve a quadratic inequality graphically.

• Step 1. Write the quadratic inequality in standard form.
• Step 2. Graph the function f ( x ) = a x 2 + b x + c . f ( x ) = a x 2 + b x + c .
• Step 3. Determine the solution from the graph.

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the x -axis but note how the position of the parabola affects the solution.

Example 9.65

Solve − x 2 − 8 x − 12 ≤ 0 − x 2 − 8 x − 12 ≤ 0 graphically. Write the solution in interval notation.

Try It 9.129

ⓐ Solve − x 2 − 6 x − 5 > 0 − x 2 − 6 x − 5 > 0 graphically and ⓑ write the solution in interval notation.

Try It 9.130

ⓐ Solve − x 2 + 10 x − 16 ≤ 0 − x 2 + 10 x − 16 ≤ 0 graphically and ⓑ write the solution in interval notation.

Solve Quadratic Inequalities Algebraically

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the critical points for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the critical points to divide the number line into intervals and then determine whether the quadratic expression willl be postive or negative in the interval. We then determine the solution for the inequality.

Example 9.66

How to solve quadratic inequalities algebraically.

Solve x 2 − x − 12 ≥ 0 x 2 − x − 12 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.131

Solve x 2 + 2 x − 8 ≥ 0 x 2 + 2 x − 8 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.132

Solve x 2 − 2 x − 15 ≤ 0 x 2 − 2 x − 15 ≤ 0 algebraically. Write the solution in interval notation.

In this example, since the expression x 2 − x − 12 x 2 − x − 12 factors nicely, we can also find the sign in each interval much like we did when we solved rational inequalities. We find the sign of each of the factors, and then the sign of the product. Our number line would like this:

The result is the same as we found using the other method.

We summarize the steps here.

Solve a quadratic inequality algebraically.

• Step 2. Determine the critical points—the solutions to the related quadratic equation.
• Step 3. Use the critical points to divide the number line into intervals.
• Step 4. Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
• Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Example 9.67

Solve - x 2 + 6 x − 7 ≥ 0 - x 2 + 6 x − 7 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.133

Solve − x 2 + 2 x + 1 ≥ 0 − x 2 + 2 x + 1 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.134

Solve − x 2 + 8 x − 14 < 0 − x 2 + 8 x − 14 < 0 algebraically. Write the solution in interval notation.

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation ax 2 + bx + c = 0. These two solutions then gave us either the two x- intercepts for the graph or the two critical points to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

For a quadratic equation of the form ax 2 + bx + c = 0, a ≠ 0 . a ≠ 0 .

The last row of the table shows us when the parabolas never intersect the x -axis. Using the Quadratic Formula to solve the quadratic equation, the radicand is a negative. We get two complex solutions.

In the next example, the quadratic inequality solutions will result from the solution of the quadratic equation being complex.

Example 9.68

Solve, writing any solution in interval notation:

ⓐ x 2 − 3 x + 4 > 0 x 2 − 3 x + 4 > 0 ⓑ x 2 − 3 x + 4 ≤ 0 x 2 − 3 x + 4 ≤ 0

We are to find the solution to x 2 − 3 x + 4 > 0 . x 2 − 3 x + 4 > 0 . Since for all values of x x the graph is above the x -axis, all values of x make the inequality true. In interval notation we write ( − ∞ , ∞ ) . ( − ∞ , ∞ ) .

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the x -axis—no part of it is below the x -axis.

We are to find the solution to x 2 − 3 x + 4 ≤ 0 . x 2 − 3 x + 4 ≤ 0 . Since for all values of x the graph is never below the x -axis, no values of x make the inequality true. There is no solution to the inequality.

Try It 9.135

Solve and write any solution in interval notation: ⓐ − x 2 + 2 x − 4 ≤ 0 − x 2 + 2 x − 4 ≤ 0 ⓑ − x 2 + 2 x − 4 ≥ 0 − x 2 + 2 x − 4 ≥ 0

Try It 9.136

Solve and write any solution in interval notation: ⓐ x 2 + 3 x + 3 < 0 x 2 + 3 x + 3 < 0 ⓑ x 2 + 3 x + 3 > 0 x 2 + 3 x + 3 > 0

Section 9.8 Exercises

Practice makes perfect.

In the following exercises, ⓐ solve graphically and ⓑ write the solution in interval notation.

x 2 + 6 x + 5 > 0 x 2 + 6 x + 5 > 0

x 2 + 4 x − 12 < 0 x 2 + 4 x − 12 < 0

x 2 + 4 x + 3 ≤ 0 x 2 + 4 x + 3 ≤ 0

x 2 − 6 x + 8 ≥ 0 x 2 − 6 x + 8 ≥ 0

− x 2 − 3 x + 18 ≤ 0 − x 2 − 3 x + 18 ≤ 0

− x 2 + 2 x + 24 < 0 − x 2 + 2 x + 24 < 0

− x 2 + x + 12 ≥ 0 − x 2 + x + 12 ≥ 0

− x 2 + 2 x + 15 > 0 − x 2 + 2 x + 15 > 0

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

x 2 + 3 x − 4 ≥ 0 x 2 + 3 x − 4 ≥ 0

x 2 + x − 6 ≤ 0 x 2 + x − 6 ≤ 0

x 2 − 7 x + 10 < 0 x 2 − 7 x + 10 < 0

x 2 − 4 x + 3 > 0 x 2 − 4 x + 3 > 0

x 2 + 8 x > − 15 x 2 + 8 x > − 15

x 2 + 8 x < − 12 x 2 + 8 x < − 12

x 2 − 4 x + 2 ≤ 0 x 2 − 4 x + 2 ≤ 0

− x 2 + 8 x − 11 < 0 − x 2 + 8 x − 11 < 0

x 2 − 10 x > − 19 x 2 − 10 x > − 19

x 2 + 6 x < − 3 x 2 + 6 x < − 3

−6 x 2 + 19 x − 10 ≥ 0 −6 x 2 + 19 x − 10 ≥ 0

−3 x 2 − 4 x + 4 ≤ 0 −3 x 2 − 4 x + 4 ≤ 0

−2 x 2 + 7 x + 4 ≥ 0 −2 x 2 + 7 x + 4 ≥ 0

2 x 2 + 5 x − 12 > 0 2 x 2 + 5 x − 12 > 0

x 2 + 3 x + 5 > 0 x 2 + 3 x + 5 > 0

x 2 − 3 x + 6 ≤ 0 x 2 − 3 x + 6 ≤ 0

− x 2 + x − 7 > 0 − x 2 + x − 7 > 0

− x 2 − 4 x − 5 < 0 − x 2 − 4 x − 5 < 0

−2 x 2 + 8 x − 10 < 0 −2 x 2 + 8 x − 10 < 0

− x 2 + 2 x − 7 ≥ 0 − x 2 + 2 x − 7 ≥ 0

Writing Exercises

Explain critical points and how they are used to solve quadratic inequalities algebraically.

Solve x 2 + 2 x ≥ 8 x 2 + 2 x ≥ 8 both graphically and algebraically. Which method do you prefer, and why?

Describe the steps needed to solve a quadratic inequality graphically.

Describe the steps needed to solve a quadratic inequality algebraically.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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Quadratic Equations and Functions

Solve Quadratic Inequalities

Learning objectives.

By the end of this section, you will be able to:

• Solve quadratic inequalities graphically
• Solve quadratic inequalities algebraically

Before you get started, take this readiness quiz.

We have learned how to solve linear inequalities and rational inequalities previously. Some of the techniques we used to solve them were the same and some were different.

We will now learn to solve inequalities that have a quadratic expression. We will use some of the techniques from solving linear and rational inequalities as well as quadratic equations.

We will solve quadratic inequalities two ways—both graphically and algebraically.

Solve Quadratic Inequalities Graphically

A quadratic equation is in standard form when written as ax 2 + bx + c = 0. If we replace the equal sign with an inequality sign, we have a quadratic inequality in standard form.

A quadratic inequality is an inequality that contains a quadratic expression.

The standard form of a quadratic inequality is written:

The graph of a quadratic function f ( x ) = ax 2 + bx + c = 0 is a parabola. When we ask when is ax 2 + bx + c < 0, we are asking when is f( x ) < 0. We want to know when the parabola is below the x -axis.

When we ask when is ax 2 + bx + c > 0, we are asking when is f ( x ) > 0. We want to know when the parabola is above the y -axis.

We list the steps to take to solve a quadratic inequality graphically.

• Write the quadratic inequality in standard form.

• Determine the solution from the graph.

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the x -axis but note how the position of the parabola affects the solution.

Solve Quadratic Inequalities Algebraically

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the critical points for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the critical points to divide the number line into intervals and then determine whether the quadratic expression willl be postive or negative in the interval. We then determine the solution for the inequality.

The result is the same as we found using the other method.

We summarize the steps here.

• Determine the critical points—the solutions to the related quadratic equation.
• Use the critical points to divide the number line into intervals.
• Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
• Determine the intervals where the inequality is correct. Write the solution in interval notation.

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation ax 2 + bx + c = 0. These two solutions then gave us either the two x- intercepts for the graph or the two critical points to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

The last row of the table shows us when the parabolas never intersect the x -axis. Using the Quadratic Formula to solve the quadratic equation, the radicand is a negative. We get two complex solutions.

In the next example, the quadratic inequality solutions will result from the solution of the quadratic equation being complex.

Solve, writing any solution in interval notation:

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the x -axis—no part of it is below the x -axis.

Solve and write any solution in interval notation:

ⓑ no solution

ⓐ no solution

Key Concepts

• Determine the critical points — the solutions to the related quadratic equation.

Section Exercises

Practice makes perfect.

In the following exercises, ⓐ solve graphically and ⓑ write the solution in interval notation.

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

no solution

Writing Exercises

Explain critical points and how they are used to solve quadratic inequalities algebraically.

Answers will vary.

Describe the steps needed to solve a quadratic inequality graphically.

Describe the steps needed to solve a quadratic inequality algebraically.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Chapter Review Exercises

Solve quadratic equations using the square root property.

Solve Quadratic Equations of the form ax 2 = k Using the Square Root Property

In the following exercises, solve using the Square Root Property.

Solve Quadratic Equations by Completing the Square

Solve Quadratic Equations Using Completing the Square

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

In the following exercises, solve by completing the square.

Solve Quadratic Equations of the form ax 2 + bx + c = 0 by Completing the Square

Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation

In the following exercises, determine the number of solutions for each quadratic equation.

ⓐ 1 ⓑ 2 ⓒ 2 ⓓ 2

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

ⓐ factor ⓑ Quadratic Formula ⓒ square root

Solve Equations in Quadratic Form

In the following exercises, solve.

Solve Applications of Quadratic Equations

Solve Applications Modeled by Quadratic Equations

In the following exercises, solve by using the method of factoring, the square root principle, or the Quadratic Formula. Round your answers to the nearest tenth, if needed.

Find two consecutive odd numbers whose product is 323.

Find two consecutive even numbers whose product is 624.

Two consecutive even numbers whose product is 624 are 24 and 26, and −24 and −26.

A triangular banner has an area of 351 square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.

Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the of the back of the case is 70 square inches. Find the height and width of the case.

The height is 14 inches and the width is 10 inches.

A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is 5 feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.

A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood’s diagonal? Round to the nearest tenth.

The length of the diagonal is 3.6 feet.

The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.

For Sophia’s graduation party, several tables of the same width will be arranged end to end to give serving table with a total area of 75 square feet. The total length of the tables will be two more than three times the width. Find the length and width of the serving table so Sophia can purchase the correct size tablecloth . Round answer to the nearest tenth.

The width of the serving table is 4.7 feet and the length is 16.1 feet.

A ball is thrown vertically in the air with a velocity of 160 ft/sec. Use the formula h = −16 t 2 + v 0 t to determine when the ball will be 384 feet from the ground. Round to the nearest tenth.

The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 360 miles. If the plane was flying at 150 mph, what was the speed of the wind that affected the plane?

The speed of the wind was 30 mph.

Ezra kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?

Two handymen can do a home repair in 2 hours if they work together. One of the men takes 3 hours more than the other man to finish the job by himself. How long does it take for each handyman to do the home repair individually?

One man takes 3 hours and the other man 6 hours to finish the repair alone.

Graph Quadratic Functions Using Properties

Recognize the Graph of a Quadratic Function

In the following exercises, graph by plotting point.

In the following exercises, determine if the following parabolas open up or down.

ⓐ up ⓑ down

Find the Axis of Symmetry and Vertex of a Parabola

In the following exercises, find ⓐ the equation of the axis of symmetry and ⓑ the vertex.

Find the Intercepts of a Parabola

In the following exercises, find the x – and y -intercepts.

In the following exercises, graph by using its properties.

Solve Maximum and Minimum Applications

In the following exercises, find the minimum or maximum value.

The maximum value is 2 when x = 2.

In the following exercises, solve. Rounding answers to the nearest tenth.

A ball is thrown upward from the ground with an initial velocity of 112 ft/sec. Use the quadratic equation h = −16 t 2 + 112 t to find how long it will take the ball to reach maximum height, and then find the maximum height.

A daycare facility is enclosing a rectangular area along the side of their building for the children to play outdoors. They need to maximize the area using 180 feet of fencing on three sides of the yard. The quadratic equation A = −2 x 2 + 180 x gives the area, A , of the yard for the length, x , of the building that will border the yard. Find the length of the building that should border the yard to maximize the area, and then find the maximum area.

The length adjacent to the building is 90 feet giving a maximum area of 4,050 square feet.

Graph Quadratic Functions Using Transformations

In the following exercises, graph each function using a vertical shift.

In the following exercises, graph each function using a horizontal shift.

In the following exercises, graph each function using transformations.

In the following exercises, graph each function.

Find a Quadratic Function from its Graph

In the following exercises, solve graphically and write the solution in interval notation.

Practice Test

Solve the following quadratic equations. Use any method.

Use the discriminant to determine the number and type of solutions of each quadratic equation.

Solve each equation.

For each parabola, find ⓐ which direction it opens, ⓑ the equation of the axis of symmetry, ⓒ the vertex, ⓓ the x- and y -intercepts, and e) the maximum or minimum value.

Graph each quadratic function using intercepts, the vertex, and the equation of the axis of symmetry.

Model the situation with a quadratic equation and solve by any method.

Find two consecutive even numbers whose product is 360.

The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.)

The diagonal is 3.8 units long.

A water balloon is launched upward at the rate of 86 ft/sec. Using the formula h = −16 t 2 + 86 t find how long it will take the balloon to reach the maximum height, and then find the maximum height. Round to the nearest tenth.

Intermediate Algebra Copyright © 2017 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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6.5 Solving Quadratic Inequalities

Learning objectives.

• Check solutions to quadratic inequalities with one variable.
• Understand the geometric relationship between solutions to quadratic inequalities and their graphs.
• Solve quadratic inequalities.

Solutions to Quadratic Inequalities

A quadratic inequality A mathematical statement that relates a quadratic expression as either less than or greater than another. is a mathematical statement that relates a quadratic expression as either less than or greater than another. Some examples of quadratic inequalities solved in this section follow.

A solution to a quadratic inequality is a real number that will produce a true statement when substituted for the variable.

Are −3, −2, and −1 solutions to x 2 − x − 6 ≤ 0 ?

Substitute the given value in for x and simplify.

x 2 − x − 6 ≤ 0 x 2 − x − 6 ≤ 0 x 2 − x − 6 ≤ 0 ( − 3 ) 2 − ( − 3 ) − 6 ≤ 0 ( − 2 ) 2 − ( − 2 ) − 6 ≤ 0 ( − 1 ) 2 − ( − 1 ) − 6 ≤ 0 9 + 3 − 6 ≤ 0 4 + 2 − 6 ≤ 0 1 + 1 − 6 ≤ 0 6 ≤ 0 ✗ 0 ≤ 0 ✓ − 4 ≤ 0 ✓

Answer: −2 and −1 are solutions and −3 is not.

Quadratic inequalities can have infinitely many solutions, one solution, or no solution. If there are infinitely many solutions, graph the solution set on a number line and/or express the solution using interval notation. Graphing the function defined by f ( x ) = x 2 − x − 6 found in the previous example we have

The result of evaluating for any x -value will be negative, zero, or positive.

f ( − 3 ) = 6 P o s i t i v e f ( x ) > 0 f ( − 2 ) = 0 Z e r o f ( x ) = 0 f ( − 1 ) = − 4 N e g a t i v e f ( x ) < 0

The values in the domain of a function that separate regions that produce positive or negative results are called critical numbers The values in the domain of a function that separate regions that produce positive or negative results. . In the case of a quadratic function, the critical numbers are the roots, sometimes called the zeros. For example, f ( x ) = x 2 − x − 6 = ( x + 2 ) ( x − 3 ) has roots −2 and 3. These values bound the regions where the function is positive (above the x -axis) or negative (below the x -axis).

Therefore x 2 − x − 6 ≤ 0 has solutions where − 2 ≤ x ≤ 3 , using interval notation [ − 2 , 3 ] . Furthermore, x 2 − x − 6 ≥ 0 has solutions where x ≤ − 2 or x ≥ 3 , using interval notation ( − ∞ , − 2 ] ∪ [ − 3 , ∞ ) .

Given the graph of f determine the solutions to f ( x ) > 0 :

From the graph we can see that the roots are −4 and 2. The graph of the function lies above the x -axis ( f ( x ) > 0 ) in between these roots.

Because of the strict inequality, the solution set is shaded with an open dot on each of the boundaries. This indicates that these critical numbers are not actually included in the solution set. This solution set can be expressed two ways,

{ x | − 4 < x < 2 } S e t   N o t a t i o n ( − 4 , 2 ) I n t e r v a l   N o t a t i o n

In this textbook, we will continue to present answers in interval notation.

Answer: ( − 4 , 2 )

Try this! Given the graph of f determine the solutions to f ( x ) < 0 :

Answer: ( − ∞ , − 4 ) ∪ ( 2 , ∞ )

Solving Quadratic Inequalities

Next we outline a technique used to solve quadratic inequalities without graphing the parabola. To do this we make use of a sign chart A model of a function using a number line and signs (+ or −) to indicate regions in the domain where the function is positive or negative. which models a function using a number line that represents the x -axis and signs (+ or −) to indicate where the function is positive or negative. For example,

The plus signs indicate that the function is positive on the region. The negative signs indicate that the function is negative on the region. The boundaries are the critical numbers, −2 and 3 in this case. Sign charts are useful when a detailed picture of the graph is not needed and are used extensively in higher level mathematics. The steps for solving a quadratic inequality with one variable are outlined in the following example.

Solve: − x 2 + 6 x + 7 ≥ 0 .

It is important to note that this quadratic inequality is in standard form, with zero on one side of the inequality.

Step 1 : Determine the critical numbers. For a quadratic inequality in standard form, the critical numbers are the roots. Therefore, set the function equal to zero and solve.

− x 2 + 6 x + 7 = 0 − ( x 2 − 6 x − 7 ) = 0 − ( x + 1 ) ( x − 7 ) = 0 x + 1 = 0 or x − 7 = 0 x = − 1 x = 7

The critical numbers are −1 and 7.

Step 2 : Create a sign chart. Since the critical numbers bound the regions where the function is positive or negative, we need only test a single value in each region. In this case the critical numbers partition the number line into three regions and we choose test values x = − 3 , x = 0 , and x = 10 .

Test values may vary. In fact, we need only determine the sign (+ or −) of the result when evaluating f ( x ) = − x 2 + 6 x + 7 = − ( x + 1 ) ( x − 7 ) . Here we evaluate using the factored form.

f ( − 3 ) = − ( − 3 + 1 ) ( − 3 − 7 ) = − ( − 2 ) ( − 10 ) = − N e g a t i v e f ( 0 ) = − ( 0 + 1 ) ( 0 − 7 ) = − ( 1 ) ( − 7 ) = + P o s i t i v e f ( 10 ) = − ( 10 + 1 ) ( 10 − 7 ) = − ( 11 ) ( 3 ) = − N e g a t i v e

Since the result of evaluating for −3 was negative, we place negative signs above the first region. The result of evaluating for 0 was positive, so we place positive signs above the middle region. Finally, the result of evaluating for 10 was negative, so we place negative signs above the last region, and the sign chart is complete.

Step 3 : Use the sign chart to answer the question. In this case, we are asked to determine where f ( x ) ≥ 0 , or where the function is positive or zero. From the sign chart we see this occurs when x -values are inclusively between −1 and 7.

Using interval notation, the shaded region is expressed as [ − 1 , 7 ] . The graph is not required; however, for the sake of completeness it is provided below.

Indeed the function is greater than or equal to zero, above or on the x -axis, for x -values in the specified interval.

Answer: [ − 1 , 7 ]

Solve: 2 x 2 − 7 x + 3 > 0 .

Begin by finding the critical numbers, in this case, the roots of f ( x ) = 2 x 2 − 7 x + 3 .

2 x 2 − 7 x + 3 = 0 ( 2 x − 1 ) ( x − 3 ) = 0 2 x − 1 = 0 or x − 3 = 0 2 x = 1 x = 3 x = 1 2

The critical numbers are 1 2 and 3. Because of the strict inequality > we will use open dots.

Next choose a test value in each region and determine the sign after evaluating f ( x ) = 2 x 2 − 7 x + 3 = ( 2 x − 1 ) ( x − 3 ) . Here we choose test values −1, 2, and 5.

f ( − 1 ) = [ 2 ( − 1 ) − 1 ] ( − 1 − 3 ) = ( − ) ( − ) = + f ( 2 ) = [ 2 ( 2 ) − 1 ] ( 2 − 3 ) = ( + ) ( − ) = − f ( 5 ) = [ 2 ( 5 ) − 1 ] ( 5 − 3 ) = ( + ) ( + ) = +

And we can complete the sign chart.

The question asks us to find the x -values that produce positive results (greater than zero). Therefore, shade in the regions with a + over them. This is the solution set.

Answer: ( − ∞ , 1 2 ) ∪ ( 3 , ∞ )

Sometimes the quadratic function does not factor. In this case we can make use of the quadratic formula.

Solve: x 2 − 2 x − 11 ≤ 0 .

Find the critical numbers.

x 2 − 2 x − 11 = 0

Identify a , b , and c for use in the quadratic formula. Here a = 1 , b = − 2 , and c = − 11 . Substitute the appropriate values into the quadratic formula and then simplify.

x = − b ± b 2 − 4 a c 2 a = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( − 11 ) 2 ( 1 ) = 2 ± 48 2 = 2 ± 4 3 2 = 1 ± 2 3

Therefore the critical numbers are 1 − 2 3 ≈ − 2.5 and 1 + 2 3 ≈ 4.5 . Use a closed dot on the number to indicate that these values will be included in the solution set.

Here we will use test values −5, 0, and 7.

f ( − 5 ) = ( − 5 ) 2 − 2 ( − 5 ) − 11 = 25 + 10 − 11 = + f ( 0 ) = ( 0 ) 2 − 2 ( 0 ) − 11 = 0 + 0 − 11 = − f ( 7 ) = ( 7 ) 2 − 2 ( 7 ) − 11 = 49 − 14 − 11 = +

After completing the sign chart shade in the values where the function is negative as indicated by the question ( f ( x ) ≤ 0 ).

Answer: [ 1 − 2 3 , 1 + 2 3 ]

Try this! Solve: 9 − x 2 > 0 .

Answer: ( − 3 , 3 )

It may be the case that there are no critical numbers.

Solve: x 2 − 2 x + 3 > 0 .

To find the critical numbers solve,

x 2 − 2 x + 3 = 0

Substitute a = 1 , b = − 2 , and c = 3 into the quadratic formula and then simplify.

x = − b ± b 2 − 4 a c 2 a = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( 3 ) 2 ( 1 ) = 2 ± − 8 2 = 2 ± 2 i 2 2 = 1 + i 2

Because the solutions are not real, we conclude there are no real roots; hence there are no critical numbers. When this is the case, the graph has no x -intercepts and is completely above or below the x -axis. We can test any value to create a sign chart. Here we choose x = 0 .

f ( 0 ) = ( 0 ) 2 − 2 ( 0 ) + 3 = +

Because the test value produced a positive result the sign chart looks as follows:

We are looking for the values where f ( x ) > 0 ; the sign chart implies that any real number for x will satisfy this condition.

Answer: ( − ∞ , ∞ )

The function in the previous example is graphed below.

We can see that it has no x -intercepts and is always above the x -axis (positive). If the question was to solve x 2 − 2 x + 3 < 0 , then the answer would have been no solution. The function is never negative.

Try this! Solve: 9 x 2 − 12 x + 4 ≤ 0 .

Answer: One solution, 2 3 .

Find the domain: f ( x ) = x 2 − 4 .

Recall that the argument of a square root function must be nonnegative. Therefore, the domain consists of all real numbers for x such that x 2 − 4 is greater than or equal to zero.

x 2 − 4 ≥ 0

It should be clear that x 2 − 4 = 0 has two solutions x = ± 2 ; these are the critical values. Choose test values in each interval and evaluate f ( x ) = x 2 − 4 .

f ( − 3 ) = ( − 3 ) 2 − 4 = 9 − 4 = + f ( 0 ) = ( 0 ) 2 − 4 = 0 − 4 = − f ( 3 ) = ( 3 ) 2 − 4 = 9 − 4 = +

Shade in the x -values that produce positive results.

Answer: Domain: ( − ∞ , − 2 ] ∪ [ 2 , ∞ )

Key Takeaways

• Quadratic inequalities can have infinitely many solutions, one solution or no solution.
• We can solve quadratic inequalities graphically by first rewriting the inequality in standard form, with zero on one side. Graph the quadratic function and determine where it is above or below the x -axis. If the inequality involves “less than,” then determine the x -values where the function is below the x -axis. If the inequality involves “greater than,” then determine the x -values where the function is above the x -axis.
• We can streamline the process of solving quadratic inequalities by making use of a sign chart. A sign chart gives us a visual reference that indicates where the function is above the x -axis using positive signs or below the x -axis using negative signs. Shade in the appropriate x -values depending on the original inequality.
• To make a sign chart, use the function and test values in each region bounded by the roots. We are only concerned if the function is positive or negative and thus a complete calculation is not necessary.

Topic Exercises

Part a: solutions to quadratic inequalities.

Determine whether or not the given value is a solution.

x 2 − x + 1 < 0 ; x = − 1

x 2 + x − 1 > 0 ; x = − 2

4 x 2 − 12 x + 9 ≤ 0 ; x = 3 2

5 x 2 − 8 x − 4 < 0 ; x = − 2 5

3 x 2 − x − 2 ≥ 0 ; x = 0

4 x 2 − x + 3 ≤ 0 ; x = − 1

2 − 4 x − x 2 < 0 ; x = 1 2

5 − 2 x − x 2 > 0 ; x = 0

− x 2 − x − 9 < 0 ; x = − 3

− x 2 + x − 6 ≥ 0 ; x = 6

Given the graph of f determine the solution set.

f ( x ) ≤ 0 ;

f ( x ) ≥ 0 ;

f ( x ) > 0 ;

f ( x ) < 0 ;

Use the transformations to graph the following and then determine the solution set.

x 2 − 1 > 0

x 2 + 2 > 0

( x − 1 ) 2 > 0

( x + 2 ) 2 ≤ 0

( x + 2 ) 2 − 1 ≤ 0

( x + 3 ) 2 − 4 > 0

− x 2 + 4 ≥ 0

− ( x + 2 ) 2 > 0

− ( x + 3 ) 2 + 1 < 0

− ( x − 4 ) 2 + 9 > 0

Part B: Solving Quadratic Inequalities

Use a sign chart to solve and graph the solution set. Present answers using interval notation.

x 2 − x − 12 > 0

x 2 − 10 x + 16 > 0

x 2 + 2 x − 24 < 0

x 2 + 15 x + 54 < 0

x 2 − 23 x − 24 ≤ 0

x 2 − 12 x + 20 ≤ 0

2 x 2 − 11 x − 6 ≥ 0

3 x 2 + 17 x − 6 ≥ 0

8 x 2 − 18 x − 5 < 0

10 x 2 + 17 x + 6 > 0

9 x 2 + 30 x + 25 ≤ 0

16 x 2 − 40 x + 25 ≤ 0

4 x 2 − 4 x + 1 > 0

9 x 2 + 12 x + 4 > 0

− x 2 − x + 30 ≥ 0

− x 2 − 6 x + 27 ≤ 0

x 2 − 64 < 0

x 2 − 81 ≥ 0

4 x 2 − 9 ≥ 0

16 x 2 − 25 < 0

25 − 4 x 2 ≥ 0

1 − 49 x 2 < 0

x 2 − 8 > 0

x 2 − 75 ≤ 0

2 x 2 + 1 > 0

4 x 2 + 3 < 0

x − x 2 > 0

3 x − x 2 ≤ 0

x 2 − x + 1 < 0

x 2 + x − 1 > 0

4 x 2 − 12 x + 9 ≤ 0

5 x 2 − 8 x − 4 < 0

3 x 2 − x − 2 ≥ 0

4 x 2 − x + 3 ≤ 0

2 − 4 x − x 2 < 0

5 − 2 x − x 2 > 0

− x 2 − x − 9 < 0

− x 2 + x − 6 ≥ 0

− 2 x 2 + 4 x − 1 ≥ 0

− 3 x 2 − x + 1 ≤ 0

Find the domain of the function.

f ( x ) = x 2 − 25

f ( x ) = x 2 + 3 x

g ( x ) = 3 x 2 − x − 2

g ( x ) = 12 x 2 − 9 x − 3

h ( x ) = 16 − x 2

h ( x ) = 3 − 2 x − x 2

f ( x ) = x 2 + 10

f ( x ) = 9 + x 2

A robotics manufacturing company has determined that its weekly profit in thousands of dollars is modeled by P ( n ) = − n 2 + 30 n − 200 where n represents the number of units it produces and sells. How many units must the company produce and sell to maintain profitability. (Hint: Profitability occurs when profit is greater than zero.)

The height in feet of a projectile shot straight into the air is given by h ( t ) = − 16 t 2 + 400 t where t represents the time in seconds after it is fired. In what time intervals is the projectile under 1,000 feet? Round to the nearest tenth of a second.

Part C: Discussion Board

Does the sign chart for any given quadratic function always alternate? Explain and illustrate your answer with some examples.

Research and discuss other methods for solving a quadratic inequality.

Explain the difference between a quadratic equation and a quadratic inequality. How can we identify and solve each? What is the geometric interpretation of each?

[ − 4 , 2 ]

[ − 1 , 3 ]

( − ∞ , ∞ )

( − ∞ , 4 ) ∪ ( 8 , ∞ )

( − ∞ , − 1 ) ∪ ( 1 , ∞ )

( − ∞ , 1 ) ∪ ( 1 , ∞ )

[ − 3 , − 1 ]

[ − 2 , 2 ]

( − ∞ , − 4 ) ∪ ( − 2 , ∞ )

( − ∞ , − 3 ) ∪ ( 4 , ∞ )

( − 6 , 4 )

[ − 1 , 24 ]

• ( − ∞ , − 1 2 ] ∪ [ 6 , ∞ )
• ( − 1 4 , 5 2 )
• ( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ )

[ − 6 , 5 ]

( − 8 , 8 )

• ( − ∞ , − 3 2 ] ∪ [ 3 2 , ∞ )
• [ − 5 2 , 5 2 ]

( − ∞ , − 2 2 ) ∪ ( 2 2 , ∞ )

• ( − ∞ , − 2 3 ] ∪ [ 1 , ∞ )

( − ∞ , − 2 − 6 ) ∪ ( − 2 + 6 , ∞ )

• [ 2 − 2 2 , 2 + 2 2 ]
• ( − ∞ , − 5 ] ∪ [ 5 , ∞ )

[ − 4 , 4 ]

The company must produce and sell more than 10 units and fewer than 20 units each week.

Answer may vary

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Algebra II : Quadratic Inequalities

Study concepts, example questions & explanations for algebra ii, all algebra ii resources, example questions, example question #1 : quadratic inequalities.

Solve the following quadratic inequality, and report your answer in interval form:

The problem is already in standard form, so all we have to at first do is set the quadratic expression = 0 and factor as normal.

Negative x^2's are hard to work with, so we multiply through by -1.

Now we can factor easily.

By the zero product property, each of these factors will be equal to 0.

Since -9 and 1 are our zeros, we just have to test one point in the region between them to find out which region our answer set goes in. Let's test x = 0 in the original inequality.

Since this statement is false, the region between -9 and 1 is not correct. So it must be the region on either side of those points. Since the original inequality was less than or equal to, the boundary points are included. So all values from -infinity to -9 inclusive, and from 1 inclusive to infinity, are solutions. In interval notation we write this as:

Example Question #2 : Quadratic Inequalities

Solve the following quadratic inequality:

First we want to rewrite the quadratic in standard form:

Now we want to set it = 0 and factor and solve like normal.

Using the zero product property, both factors produce a zero:

So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality.

Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.

Example Question #3 : Quadratic Inequalities

1. Rewrite the equation in standard form.

Example Question #4 : Quadratic Inequalities

Solve this inequality.

Combine like terms first.

The zeroes are 3 and 8 so a number line can be divided into 3 sections.

X<3 works, 3<x<8 does not work, and x>8 works

Example Question #5 : Quadratic Inequalities

Now, plot these two points on to a number line.

Notice that these two numbers effectively divide up the number line into three regions:

Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true.

Since this will make the inequality true, the solution can lie in this region.

Since this number is not less than zero, the solution cannot lie in this region.

Example Question #7 : Quadratic Inequalities

The solution cannot be determined with the information given.

Now, plot these two numbers on to a number line.

Notice how these numbers divide the number line into three regions:

Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.

Since this is not less than zero, the solution to the inequality cannot be found in this region.

Since this is less than zero, the solution is found in this region.

Since this is not less than zero, the solution is not found in this region.

Example Question #6 : Quadratic Inequalities

Now, plot these two numbers on a number line.

Notice how the number line is divided into three regions:

Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.

Since this number is not less than zero, the solution cannot be found in this region.

Since this number is less than zero, the solution can be found in this region.

Example Question #10 : Quadratic Inequalities

Not enough information to solve

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Course: Algebra 2   >   Unit 12

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Number Line

• \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
• \mathrm{Kira\:went\:for\:a\:drive\:in\:her\:new\:car.\:She\:drove\:for\:142.5\:miles\:at\:a\:speed\:of\:57\:mph.\:For\:how\:many\:hours\:did\:she\:drive?}
• \mathrm{The\:sum\:of\:two\:numbers\:is\:249\:.\:Twice\:the\:larger\:number\:plus\:three\:times\:the\:smaller\:number\:is\:591\:.\:Find\:the\:numbers.}
• \mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
• \mathrm{You\:deposit\:3000\:in\:an\:account\:earning\:2\%\:interest\:compounded\:monthly.\:How\:much\:will\:you\:have\:in\:the\:account\:in\:15\:years?}
• How do you solve word problems?
• To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
• How do you identify word problems in math?
• Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
• Is there a calculator that can solve word problems?
• Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
• What is an age problem?
• An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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• High School Math Solutions – Inequalities Calculator, Exponential Inequalities Last post, we talked about how to solve logarithmic inequalities. This post, we will learn how to solve exponential...

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Condensed Matter > Statistical Mechanics

Title: annealing-assisted column generation for inequality-constrained combinatorial optimization problems.

Abstract: Ising machines are expected to solve combinatorial optimization problems faster than the existing integer programming solvers. These problems, particularly those encountered in practical situations, typically involve inequality constraints. However, owing to the hardware limitations of the current Ising machines, solving combinatorial optimization problems with inequality constraints remains challenging. The Capacitated Vehicle Routing Problem (CVRP) is a typical example of a problem with inequality constraints. The objective function of the CVRP is to minimize the total distance traveled by each vehicle while limiting the total demand of customers served by a single vehicle to the vehicle's capacity. The CVRP is classified as NP-hard and, thus, is commonly solved using heuristic algorithms, such as column generation. Column generation attempts to iteratively generate only the promising routes, as the number of feasible routes increases exponentially. Within this framework, the CVRP is formulated as a set cover problem. The corresponding dual solutions are used to define the pricing subproblem, which is intended to create a new route. By applying Ising machines to this pricing subproblem, the overall computation time can be reduced. This study aims to solve combinatorial optimization problems with inequality constraints using a hybrid algorithm that combines column generation and Ising machines, thereby extending the applications of the latter. We parameterize the difficulty of the inequality constraints and demonstrate that our annealing-assisted column generation can converge to a better lower bound.

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