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Difficulty Level: At Grade Created by: CK-12
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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?

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### Guidance

Quadratic inequalities are inequalities that have one of the following forms

ax2+bx+c>0\begin{align*}ax^{2}+bx+c>0\end{align*}

and

ax2+bx+c<0\begin{align*}ax^{2}+bx+c<0\end{align*}

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)=x2+x6\begin{align*}y=f(x)=x^{2}+x-6\end{align*}

Notice that the curve intersects the x\begin{align*}x-\end{align*}axis at -3 and 2. From graph, we notice the followings

• If x<3\begin{align*}x< -3\end{align*} then f(x)>0\begin{align*}f(x)>0\end{align*}
• If 3<x<2\begin{align*}-3, then f(x)<0\begin{align*}f(x)<0\end{align*}
• If x>2\begin{align*}x>2\end{align*}, then f(x)>0\begin{align*}f(x)>0\end{align*}

Therefore, x2+x6>0\begin{align*}x^{2}+x-6>0\end{align*} whenever x<3\begin{align*}x<-3\end{align*} or x>2\begin{align*}x>2\end{align*}, and x2+x6<0\begin{align*}x^{2}+x-6<0\end{align*} when 3<x<2\begin{align*}-3.

#### Example A

What is the solution set of the inequality 2x2+7x4<0\begin{align*}2x^{2}+7x-4<0\end{align*}?

Solution

It is best to graph the function f(x)=2x2+7x4\begin{align*}f(x)=2x^{2}+7x-4\end{align*} and look for the the values of x\begin{align*}x\end{align*} such that the inequality f(x)<0\begin{align*}f(x)<0\end{align*} is true.

Thus from graph, 2x2+7x4<0\begin{align*}2x^{2}+7x-4<0\end{align*} only if

4<x<12\begin{align*}-4

So the solution set is x(4,12)\begin{align*}x\in\left ( -4,\frac{1}{2} \right )\end{align*} or in set builder notation, {x|4<x<12}\begin{align*}\left \{x|-4.

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\begin{align*}x-\end{align*}intercepts of the graph and then dividing the x\begin{align*}x-\end{align*}axis into intervals separated by the x\begin{align*}x-\end{align*}intercepts. The examples below illustrate the method.

#### Example B

Find the solution set of the quadratic inequality x2+2x8>0\begin{align*}x^{2}+2x-8>0\end{align*} without graphing.

Solution

To find the solution set without graphing, first factor:

x2+2x8=0\begin{align*}x^2 + 2x - 8 = 0\end{align*}
(x+4)(x2)=0\begin{align*}(x + 4)(x - 2) = 0\end{align*}

Recalling the zero product rule, we can see that the two solutions to this quadratic equation are x=4\begin{align*}x = -4\end{align*} and x=2\begin{align*}x = 2\end{align*}, thus, the x\begin{align*}x-\end{align*}intercepts of the function f(x)=x2+2x8\begin{align*}f(x) = x^{2} + 2x - 8\end{align*} are -4 and 2.

These points divide the x\begin{align*}x-\end{align*}axis into three intervals: (,4)|(4,2)|(2,)\begin{align*}(-\infty,-4) | (-4,2) | (2,\infty)\end{align*}. We can choose a test point from each interval, substitute it into f(x)\begin{align*}f(x)\end{align*} 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 x2+2x8\begin{align*}x^{2}+2x-8\end{align*} positive or negative? Part of Solution set?
(,4)\begin{align*}(-\infty,-4)\end{align*} 5\begin{align*}-5\end{align*} +\begin{align*}+\end{align*} yes\begin{align*}yes\end{align*}
(4,2)\begin{align*}(-4, 2)\end{align*} 1\begin{align*}1\end{align*} \begin{align*}-\end{align*} no\begin{align*}no\end{align*}
(2,+)\begin{align*}(2,+\infty)\end{align*} 3\begin{align*}3\end{align*} +\begin{align*}+\end{align*} yes\begin{align*}yes\end{align*}

From the table, we conclude that since x2+2x8>0\begin{align*}x^2 +2x -8 > 0\end{align*} if and only if x<4\begin{align*}x < -4\end{align*} and x>2\begin{align*}x > 2\end{align*}. The solution set can also be written as:

x(,4)(2,+)\begin{align*}x\in(-\infty,-4)\cup(2,+\infty)\end{align*}

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\begin{align*}w\end{align*} be the width of the rectangle and l\begin{align*}l\end{align*} its length. Given the information in the question, we can say:

l=10+2w\begin{align*}l = 10 + 2w\end{align*}

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

area=l×w\begin{align*}area = l \times w\end{align*}

Substituting 10+2w\begin{align*}10 + 2w\end{align*} in for l\begin{align*}l\end{align*} gives:

area=w×(10+2w)\begin{align*}area = w \times (10 + 2w)\end{align*}
=2w2+10w\begin{align*}=2w^2 + 10w\end{align*}

The area can not exceed 100 m2\begin{align*}100 \ m^{2}\end{align*} so

10w+2w2<100\begin{align*}10w+2w^{2}<100\end{align*}

or

2w2+10w100<0\begin{align*}2w^{2}+10w-100<0\end{align*}

Simplify by dividing both sides by 2:

w2+5w50<0\begin{align*}w^{2}+5w-50<0\end{align*}

Factor the trinomial:

w2+5w50=(w+10)(w5)\begin{align*}w^{2}+5w-50=(w+10)(w-5)\end{align*}

So the partition points are 5 and -10, which means we have three intervals. Since width can not be negative, we can safely ignore -10. That means the maximum area is 100 m2\begin{align*}100 \ m^{2}\end{align*}

w<5\begin{align*}\therefore w < 5\end{align*}.
The width must be less than 5 meters.

Concept question follow-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 S2<6250\begin{align*}S^{2} < 6250\end{align*} describes the possible side lengths of square shapes you could mow before running out of gas.

Solving for S gives:

s2=6250\begin{align*}\sqrt{s^2} = \sqrt{6250}\end{align*}
s=79\begin{align*}s = 79\end{align*}

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 x216\begin{align*}x^2 \leq 16\end{align*}

2) Find the solution set: x27x>30\begin{align*}x^2 - 7x > 30\end{align*}

3) Find the solution set: x213x>36\begin{align*}x^2 - 13x > -36\end{align*}

4) Graph the solution set: (x3)(x+4)0\begin{align*}(x - 3)(x + 4) \geq 0\end{align*}

Solutions

1) Set the function equal to zero:

x216=0\begin{align*}x^2 - 16 = 0\end{align*}
Factor to find the critical values (points where the graph crosses the x axis, thereby changing signs):
(x4)(x+4)=0\begin{align*}(x - 4)(x + 4) = 0\end{align*}
By the zero product rule: x=4\begin{align*}x = 4\end{align*} or x=4\begin{align*}x = -4\end{align*}
That gives us three sections on the graph:
x4\begin{align*}x \leq -4\end{align*}
4x4\begin{align*}-4 \leq x \leq 4\end{align*}
4x\begin{align*}4 \leq x\end{align*}
Test one sample value from each division to identify possible solution sets.
Set x4\begin{align*}x \leq -4\end{align*} 4x4\begin{align*}-4 \leq x \leq 4\end{align*} 4x\begin{align*}4 \leq x\end{align*}
Test value 5\begin{align*}-5\end{align*} 0\begin{align*}0\end{align*} 5\begin{align*}5\end{align*}
f(x)\begin{align*}f(x)\end{align*} true with value? 4(5)2=10016\begin{align*}4\cdot(-5)^2 = 100 \nleq 16\end{align*} No\begin{align*}\therefore No\end{align*} 402=016\begin{align*}4\cdot 0^2 = 0 \leq 16\end{align*} Yes\begin{align*}\therefore Yes\end{align*} 452=10016\begin{align*}4\cdot5^2 = 100 \nleq 16\end{align*} No\begin{align*}\therefore No\end{align*}
Therefore the solution set is 4x4\begin{align*}- 4 \leq x \leq 4\end{align*}

2) Follow the same process as #1:

Set the function equal to zero: x27x30=0\begin{align*}x^2 - 7x - 30 = 0\end{align*}
Factor: (x10)(x+3)=0\begin{align*}(x - 10)(x + 3) = 0\end{align*}
Identify critical values: x=3,10\begin{align*}x = { -3, 10 }\end{align*}
The three sections are: x<3\begin{align*}x < -3\end{align*} and 3<x<10\begin{align*}-3 < x < 10\end{align*} and 10<x\begin{align*}10 < x\end{align*}
Test one sample value from each division to identify possible solution sets.
Set x<3\begin{align*}x < -3\end{align*} 3<x<10\begin{align*}-3 < x < 10\end{align*} 10<x\begin{align*}10 < x\end{align*}
Test value 10\begin{align*}-10\end{align*} 0\begin{align*}0\end{align*} 20\begin{align*}20\end{align*}
f(x)\begin{align*}f(x)\end{align*} true with value? (10)27(10)=170\begin{align*}(-10)^2 -7(-10) = 170\end{align*} 170>30\begin{align*}170 > 30\end{align*} Yes\begin{align*}\therefore Yes\end{align*} 027(0)=0\begin{align*}0^2 -7(0) = 0\end{align*} 0<30\begin{align*}0 < 30\end{align*} No\begin{align*}\therefore No\end{align*} 2027(20)=260\begin{align*}20^2 -7(20) = 260\end{align*} 260>30\begin{align*}260 > 30\end{align*} Yes\begin{align*}\therefore Yes\end{align*}
The solution set is: x<3\begin{align*}x < -3\end{align*} and 10<x\begin{align*}10 < x\end{align*}

3) Use the same process again:

Set the function equal to zero: x213x+36=0\begin{align*}x^2 - 13x + 36 = 0\end{align*}
Factor: (x9)(x4)=0\begin{align*}(x - 9)(x - 4) = 0\end{align*}
Identify critical values: x=4,9\begin{align*}x = {4, 9}\end{align*}
The three sections are: x<4\begin{align*}x < 4\end{align*} and 4<x<9\begin{align*}4 < x < 9\end{align*} and 9<x\begin{align*}9 < x\end{align*}
Test one sample value from each division to identify possible solution sets.
Set x<4\begin{align*}x < 4\end{align*} 4<x<9\begin{align*}4 < x < 9\end{align*} 9<x\begin{align*}9 < x\end{align*}
Test value 0\begin{align*}0\end{align*} 6\begin{align*}6\end{align*} 12\begin{align*}12\end{align*}
f(x)\begin{align*}f(x)\end{align*} true with value? (0)213(0)=0\begin{align*}(0)^2 -13(0) = 0\end{align*} 0>36\begin{align*}0 > -36\end{align*} Yes\begin{align*}\therefore Yes\end{align*} 6213(6)=42\begin{align*}6^2 -13(6) = -42\end{align*} 42<36\begin{align*}-42 < -36\end{align*} No\begin{align*}\therefore No\end{align*} 12213(12)=12\begin{align*}12^2 -13(12) = -12\end{align*} 12>36\begin{align*}-12 > -36\end{align*} Yes\begin{align*}\therefore Yes\end{align*}
The solution set is: x<4\begin{align*}x < 4\end{align*} and 9<x\begin{align*}9 < x\end{align*}

4) The solutions to (x3)(x+4)0\begin{align*}(x - 3)(x + 4) \geq 0\end{align*} 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 (x3)(x+4)0\begin{align*}(x - 3)(x + 4) \geq 0\end{align*} we know we need a positive answer or zero.
Therefore either:
Case #1: (x3)0x3\begin{align*}(x - 3) \geq 0 \to x \geq 3\end{align*} and (x+4)0x4\begin{align*}(x + 4) \geq 0 \to x \geq -4\end{align*}
or
Case #2: (x3)0x3\begin{align*}(x - 3) \leq 0 \to x \leq 3\end{align*} and (x+4)0x4\begin{align*}(x + 4) \leq 0 \to x \leq -4\end{align*}
Since any number greater than 3 is already greater than -4, from Case #1 we get: x3\begin{align*}x \geq 3\end{align*}
Since any number less than -4 is already less than 3, from Case #2 we get x4\begin{align*}x \leq -4\end{align*}
Therefore our answer is x4\begin{align*}x \leq -4\end{align*} or x3\begin{align*}x \geq 3\end{align*}
In set notation: x(,4][3,+)\begin{align*}x\in(-\infty,-4]\cup[3,+\infty)\end{align*}
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:

### Practice

Graph the solutions sets below on a number line:

1. x<3\begin{align*}x < 3\end{align*} or x>4\begin{align*}x > 4\end{align*}
2. x5\begin{align*}x \geq -5\end{align*} and x3\begin{align*}x \geq 3\end{align*}
3. x<6\begin{align*}x < 6\end{align*} and x2\begin{align*}x \geq -2\end{align*}
4. x>7\begin{align*}x > 7\end{align*} or x4\begin{align*}x \geq -4\end{align*}
5. x8\begin{align*}x \leq -8\end{align*} and x>3\begin{align*}x > 3\end{align*}

Identify Critical points, solve, and graph:

1. x2+9x>14\begin{align*}x^2 + 9x > -14\end{align*}
2. x25x50\begin{align*}x^2 -5x \leq 50\end{align*}
3. x2+2x48\begin{align*}x^2 + 2x \leq 48\end{align*}
4. x20x8<0\begin{align*}x - \frac{20}{x} - 8 < 0\end{align*} (hint: multiply both sides by x first)
5. x+1021x\begin{align*}x + 10 \geq -\frac{21}{x}\end{align*}
6. (x+6)(x3)>0\begin{align*}(x + 6)(x - 3) > 0\end{align*}
7. (x8)(x+1)>0\begin{align*}(x - 8)(x + 1) > 0\end{align*}
8. x2x90\begin{align*}x^2 - x \geq 90\end{align*}
9. 3x223x8\begin{align*}3x^2 - 23x \leq 8\end{align*}
10. x2+x6>0\begin{align*}x^2 + x - 6 > 0\end{align*}

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Color Highlighted Text Notes

### Vocabulary Language: English

Critical Values

The critical values are the $x-$intercepts of a quadratic function.

interval

An interval is a specific and limited part of a function.

A quadratic inequality is a quadratic expression that is specified to be greater or lesser than a given value.

Roots

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

Zero Product Rule

The zero product rule states that if the product of two expressions is equal to zero, then at least one of the original expressions much be equal to zero.

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