<meta http-equiv="refresh" content="1; url=/nojavascript/"> Quadratic Inequalities ( Read ) | Analysis | CK-12 Foundation

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How would you express the following as a function?

You are supposed to mow your square-shaped lawn for your parents, but the mower only has part of a tank of gas. If you can mow 2500 sf per gallon, and the mower has approximately 2.5 gallons in it, what is the maximum length of one side of the lawn you can mow? If your lawn is 75 feet long, will you need more gas?

### Guidance

Quadratic inequalities are inequalities that have one of the following forms

$ax^{2}+bx+c>0$

and

$ax^{2}+bx+c<0$

We can solve these inequalities by using the techniques that we have learned about solving quadratic equations. For example, consider the graph of the equation:

$y=f(x)=x^{2}+x-6$

Notice that the curve intersects the $x-$ axis at -3 and 2. From graph, we notice the followings

• If x<3$x< -3$ then f(x)>0$f(x)>0$
• If 3<x<2$-3 < x < 2$ , then f(x)<0$f(x)<0$
• If x>2$x>2$ , then f(x)>0$f(x)>0$

Therefore, $x^{2}+x-6>0$ whenever $x<-3$ or $x>2$ , and $x^{2}+x-6<0$ when $-3< x < 2$ .

#### Example A

What is the solution set of the inequality $2x^{2}+7x-4<0$ ?

Solution:

It is best to graph the function $f(x)=2x^{2}+7x-4$ and look for the values of $x$ such that the inequality $f(x)<0$ is true.

Thus from graph, $2x^{2}+7x-4<0$ only if

$-4 < x < \frac{1}{2}$

So the solution set is $x\in\left ( -4,\frac{1}{2} \right )$ or in set builder notation, $\left \{x | -4 < x < \frac{1}{2} \right \}$ .

Although the method of graphing to find the solution set of an inequality is easy to follow, another algebraic method can be used. The algebraic method involves finding the $x-$ intercepts of the graph and then dividing the $x-$ axis into intervals separated by the $x-$ intercepts. The examples below illustrate the method.

#### Example B

Find the solution set of the quadratic inequality $x^{2}+2x-8>0$ without graphing.

Solution:

To find the solution set without graphing, first factor:

$x^2 + 2x - 8 = 0$
$(x + 4)(x - 2) = 0$

Recalling the zero product rule , we can see that the two solutions to this quadratic equation are $x = -4$ and $x = 2$ , thus, the $x-$ intercepts of the function $f(x) = x^{2} + 2x - 8$ are -4 and 2.

These points divide the $x-$ axis into three intervals: $(-\infty,-4) | (-4,2) | (2,\infty)$ . We can choose a test point from each interval, substitute it into $f(x)$ and see if the function is negative or positive with that value as x . This procedure can be simplified by making a table as shown below:

Interval Test Point Is $x^{2}+2x-8$ positive or negative? Part of Solution set?
$(-\infty,-4)$ $-5$ $+$ $yes$
$(-4, 2)$ $1$ $-$ $no$
$(2,+\infty)$ $3$ $+$ $yes$

From the table, we conclude that since $x^2 +2x -8 > 0$ if and only if $x < -4$ and $x > 2$ . The solution set can also be written as:

$x\in(-\infty,-4)\cup(2,+\infty)$

Some problems in science involve quadratic inequalities. The example below illustrates one such application.

#### Example C

A rectangle has a length 10 meters more than twice the width. Find all of the possible widths that result in the area of the rectangle not exceeding 100 squared meters.

Solution:

Let $w$ be the width of the rectangle and $l$ its length. Given the information in the question, we can say:

$l = 10 + 2w$

Then we can use the formula for the area of a rectangle:

$area = l \times w$

Substituting $10 + 2w$ in for $l$ gives:

$area = w \times (10 + 2w)$
$=2w^2 + 10w$

The area cannot exceed $100 \ m^{2}$ so

$10w+2w^{2}<100$

or

$2w^{2}+10w-100<0$

Simplify by dividing both sides by 2:

$w^{2}+5w-50<0$

Factor the trinomial:

$w^{2}+5w-50=(w+10)(w-5)$

So the partition points are 5 and -10, which means we have three intervals. Since width cannot be negative, we can safely ignore -10. That means the maximum area is $100 \ m^{2}$

$\therefore w < 5$ .
The width must be less than 5 meters.

Concept question wrap-up:

Were you able to solve the question about mowing a lawn that was discussed at the beginning of the lesson?

'If you can mow 2500sf of grass per gallon of gas, and the mower has 2.5 gallons in it, what is the maximum length of one side of the lawn you can mow?

If your lawn is 75ft long, will you need more gas?'

By applying the process from Ex#3, we know that the function $S^{2} < 6250$ describes the possible side lengths of square shapes you could mow before running out of gas.

Solving for S gives:

$\sqrt{s^2} = \sqrt{6250}$
$s = 79$

With 2.5gal of gas, you could mow a square up to apx 79ft on each side.

You should not need more gas if the lawn is only 75ft long on each side.

### Vocabulary

Quadratic Inequality: A term describing a squared function that is specified to be smaller or larger than a given value.

Roots: The roots of a quadratic function are the values of x that make y equal to zero.

### Guided Practice

1) Find the solution set of the inequality $x^2 \leq 16$

2) Find the solution set: $x^2 - 7x > 30$

3) Find the solution set: $x^2 - 13x > -36$

4) Graph the solution set: $(x - 3)(x + 4) \geq 0$

1) Set the function equal to zero:

$x^2 - 16 = 0$
Factor to find the critical values (points where the graph crosses the x axis, thereby changing signs):
$(x - 4)(x + 4) = 0$
By the zero product rule: $x = 4$ or $x = -4$
That gives us three sections on the graph:
$x \leq -4$
$-4 \leq x \leq 4$
$4 \leq x$
Test one sample value from each division to identify possible solution sets.
Set $x \leq -4$ $-4 \leq x \leq 4$ $4 \leq x$
Test value $-5$ $0$ $5$
$f(x)$ true with value? $4\cdot(-5)^2 = 100 \nleq 16$ $\therefore No$ $4\cdot 0^2 = 0 \leq 16$ $\therefore Yes$ $4\cdot5^2 = 100 \nleq 16$ $\therefore No$
Therefore the solution set is $- 4 \leq x \leq 4$

2) Follow the same process as #1:

Set the function equal to zero: $x^2 - 7x - 30 = 0$
Factor: $(x - 10)(x + 3) = 0$
Identify critical values: $x = { -3, 10 }$
The three sections are: $x < -3$ and $-3 < x < 10$ and $10 < x$
Test one sample value from each division to identify possible solution sets.
Set $x < -3$ $-3 < x < 10$ $10 < x$
Test value $-10$ $0$ $20$
$f(x)$ true with value? $(-10)^2 -7(-10) = 170$ $170 > 30$ $\therefore Yes$ $0^2 -7(0) = 0$ $0 < 30$ $\therefore No$ $20^2 -7(20) = 260$ $260 > 30$ $\therefore Yes$
The solution set is: $x < -3$ and $10 < x$

3) Use the same process again:

Set the function equal to zero: $x^2 - 13x + 36 = 0$
Factor: $(x - 9)(x - 4) = 0$
Identify critical values: $x = {4, 9}$
The three sections are: $x < 4$ and $4 < x < 9$ and $9 < x$
Test one sample value from each division to identify possible solution sets.
Set $x < 4$ $4 < x < 9$ $9 < x$
Test value $0$ $6$ $12$
$f(x)$ true with value? $(0)^2 -13(0) = 0$ $0 > -36$ $\therefore Yes$ $6^2 -13(6) = -42$ $-42 < -36$ $\therefore No$ $12^2 -13(12) = -12$ $-12 > -36$ $\therefore Yes$
The solution set is: $x < 4$ and $9 < x$

4) The solutions to $(x - 3)(x + 4) \geq 0$ can be identified with the rules for multiplying negative numbers:

Recall from Pre-Algebra that an even number of negatives yields a positive answer, and an odd number of negatives yields a negative answer.
Since $(x - 3)(x + 4) \geq 0$ we know we need a positive answer or zero.
Therefore either:
Case #1: $(x - 3) \geq 0 \to x \geq 3$ and $(x + 4) \geq 0 \to x \geq -4$
or
Case #2: $(x - 3) \leq 0 \to x \leq 3$ and $(x + 4) \leq 0 \to x \leq -4$
Since any number greater than 3 is already greater than -4, from Case #1 we get: $x \geq 3$
Since any number less than -4 is already less than 3, from Case #2 we get $x \leq -4$
Therefore our answer is $x \leq -4$ or $x \geq 3$
In set notation: $x\in(-\infty,-4]\cup[3,+\infty)$
To graph this information, we draw a line graph, and mark the values that x can be , with solid dots on the end numbers to indicate that those values are included.
Visually that is:

### Explore More

Graph the solutions sets below on a number line:

1. $x < 3$ or $x > 4$
2. $x \geq -5$ and $x \geq 3$
3. $x < 6$ and $x \geq -2$
4. $x > 7$ or $x \geq -4$
5. $x \leq -8$ and $x > 3$

Identify critical points, solve, and graph:

1. $x^2 + 9x > -14$
2. $x^2 -5x \leq 50$
3. $x^2 + 2x \leq 48$
4. $x - \frac{20}{x} - 8 < 0$ (hint: multiply both sides by x first)
5. $x + 10 \geq -\frac{21}{x}$
6. $(x + 6)(x - 3) > 0$
7. $(x - 8)(x + 1) > 0$
8. $x^2 - x \geq 90$
9. $3x^2 - 23x \leq 8$
10. $x^2 + x - 6 > 0$