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3. Quadratic Functions & Equations
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Algebra 2

3.04 Quadratic inequalities

Lesson
Practice
Lesson

Concept summary

We can have quadratic inequalities in one or two variables.

Quadratic inequality in one variable

An inequality involving one variable that contains a term of degree 2, but no term of higher degree. The solution is a set of values that can be shown on a number line.

Example:

2x^2>18

We can solve quadratic inequalities in one variable by solving the related equation and then determining which of the intervals on the number line represent the solutions.

The end points will be unfilled if the inequality is < or > and filled if the inequality is \leq or \geq. The interval(s) that will be included depend on the direction of the inequality. For example, x^2<9 gives:

-5-4-3-2-1012345

while x^2 \geq 9 gives:

-5-4-3-2-1012345

We can visualize where these solution sets come from by considering the graph of the corresponding quadratic equation. For example:

-5
-4
-3
-2
-1
1
2
3
4
5
x
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
2
y

We can rewrite x^2<9 as x^2-9<0 to get the corresponding quadratic equation y=x^2-9.

The inequality is true when the quadratic expression is less than 0, so the solutions will be when the graph is below the x-axis.

The solution set will be between x=3 and x=-3. The inequality symbol tells us that the end points are not included in the solution.

-5
-4
-3
-2
-1
1
2
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x
-10
-9
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-7
-6
-5
-4
-3
-2
-1
1
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y

We can rewrite x^2\geq9 as x^2-9\geq0 to get the same corresponding quadratic equation, but with a different inequality symbol.

This time the inequality is true when the quadratic expression is greater than or equal to 0.

This means that the solution set will be the values of x when the parabola is on or above the x-axis, which is x\geq3 and x\leq -3.

Notice that the end points of the solution set are the x-intercepts of the corresponding quadratic equation.

Another method for finding the solution set is to change the inequality symbol into an equals symbol to get the related equation. We can then solve the related equation to find our end points, then test values on either side of the end points to find where the inequality is true.

For example, with the equality x^2<9 we can consider the equation x^2=9. We can solve this to get the end points x=3 and x=-3. Plotting these points on the number line shows us that there are three regions we need to test.

-5-4-3-2-1012345

We can test each region by substituting a value from the region into the inequality and checking if the result is true or false.

  • (4)^2<9 is false, so we know that the region greater than 3 is not part of the solution set.

  • (0)^2<9 is true, so we know that the region between 3 and -3 is part of the solution set.

  • (-4)^2<9 is false, so we know that the region less than -3 is not part of the solution set.

Therefore the solution set for the quadratic inequality x^2<9 is:

-5-4-3-2-1012345

Worked examples

Example 1

Graph the inequality x^2-2x-15<0 on a number line.

Approach

To graph the inequality on a number line we first want to solve the related equation for x. We can then plot the solution on the number line.

Solution

We first want to solve f\left(x\right)=0, where f\left(x\right)=x^2-2x-15.

\displaystyle x^2-2x-15\displaystyle =\displaystyle 0
\displaystyle \left(x-5\right)\left(x+3\right)\displaystyle =\displaystyle 0Factor the quadratic

This gives us solutions of x=5, x=-3.

To solve the inequality \left(x-5\right)\left(x+3\right)<0 we now want to identify the regions where this is true.

We can set up a table to identify whether each factor will be positive or negative in the intervals less than, greater than, and between our two solutions of x=5 and x=-3. We can fill out the table by testing any value in each interval.

x<-3x=-3-3<x<5x=5x>5
\left(x-5\right)-0-0+
\left(x+3\right)-0+0+
\left(x-5\right)\left(x+3\right)+0-0+

We can see that \left(x-5\right)\left(x+3\right)<0 when -3<x<5. We can now graph the solution on a number line

-5-4-3-2-1012345678910

Reflection

Notice that we cannot solve this in the same way we would solve the equivalent quadratic function. That is, we cannot set both factors to be less than zero and solve them independently. This method would give us the incorrect solution of x<5, x<-3.

Example 2

Graph the corresponding quadratic function and solve each of the following inequalities.

a

x^2-4 \leq 0

Approach

The corresponding quadratic equation to this inequality is y=x^2-4.

Solution

The graph of y=x^2-4 is:

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y

The solution set to the inequality x^2-4 \leq 0 is equivalent to when y \leq 0 on the graph, which occurs on and below the x-axis.

From the graph, we can see that the end points of the solution set for the inequality will be x=-2\text{ and }x=2so the solution set for x^2-4 \leq 0 is: -2\leq x\leq 2

Reflection

It is a good idea to check our answer by substituting in an x-value that is inside the solution set to make sure that it satisfies the inequality. For example, x=0 is inside the solution set, so we can check that:

\displaystyle 0\displaystyle \geq\displaystyle x^2-4
\displaystyle 0\displaystyle \geq\displaystyle 0^2-4
\displaystyle 0\displaystyle \geq\displaystyle -4

It does satisfy the inequality, so we have chosen the correct interval for our solution set.

b

0 < \left(x-5\right)\left(x+1\right)

Approach

The corresponding quadratic equation to this inequality is y=(x-5)(x+1).

Solution

The graph of y=(x-5)(x+1) is:

-2
-1
1
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x
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
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y

The solution set to the inequality 0< (x-5)(x+1) is equivalent to when 0< y on the graph.

From the graph, we can see that the end points of the solution set for the inequality will be x=-1 \text{ and }x=5 and the solution set to 0< (x-5)(x+1) is : x<-1 \text{ or }x>5

Reflection

Notice that the type of inequality symbol (inclusive or exclusive) is the same for the inequality and the solution set. This is because it tells us whether the end points of the solution set are included or not.

Example 3

Write a corresponding quadratic inequality for the given solution set on the number line.

-5-4-3-2-1012345

Approach

The end points of the solution set are the solutions to the related equation, so we can use them to write the related equation in factored form.

Since the end points are included in the solution set, we know that the inequality symbol will be either \leq or \geq.

We can then substitute a value from the solution set into the equation and replace the equals symbol with the inequality symbol that would make the inequality true.

Solution

Since the end points of the solution set are -3 and 1, the related equation will be:

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

The value x=0 is within the solution set, so we can substitute this value into the equation to get (0+3)(0-1)=0.

We can see that (0+3)(0-1)\geq0 is false, and that (0+3)(0-1)\leq0 is true.

Therefore, a corresponding quadratic inequality for the solution set is:(x+3)(x-1)\leq0

Reflection

Since the inequality has 0 on the right-hand side, we could multiply the quadratic expression by any positive scale factor and still get the same result.

However, multiplying by a negative scale factor would require us to reverse the direction of the inequality symbol in order to make the solution set true.

Outcomes

A2.N.Q.A.1

Use units as a way to understand real-world problems.*

A2.A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems in a real-world context.*

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP2

Reason abstractly and quantitatively.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP4

Model with mathematics.

A2.MP5

Use appropriate tools strategically.

A2.MP6

Attend to precision.

A2.MP7

Look for and make use of structure.

A2.MP8

Look for and express regularity in repeated reasoning.

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