What relationships do you notice between the graph of the function and the solutions of the corresponding inequalities?
How can we use the graph to find the solutions -x^{2}+4 \lt c where c is any real number?

The solution set of a quadratic inequality are the values that make the inequality true. Similar to quadratic equations, we can visualize where these solution sets come from by considering the graph of the corresponding quadratic function.

\displaystyle x^{2}\gt4

\bm{y=x^{2}}

Corresponding function

-4

-3

-2

-1

In this inequality, we are looking for the x-values that make the y-values greater than 4, as shown by the green sections of the graph.

The blue sections of the graph are less than 4, so those values do not satisfy the inequality.

We can show the solution set on a number line, as shown below to only indicate the values of x that satisfy the inequality.

Notice that the end points at -2 and 2 are unfilled. This is because those values do not satisfy the inequality: \left(-2\right)^{2}\ngtr 4 and \left(2\right)^{2} \ngtr 4.

-4

-3

-2

-1

-4

-3

-2

-1

Another method to solve x^{2}\gt4 is by making one side of this inequality zero and then solving the equivalent inequality that results:x^{2}-4\gt0

By making one side of the inequality zero, we can now use the x-intercepts of y=x^{2}-4 as the boundary points and the solution set is where the graph lies above the x-axis, that is, the y-values are above zero.

If we change the inequality sign, the solution set will change as well. Consider the inequality x^{2}-4\leq0. Now, the blue region of the graph above will be the solution set because those are the values that are below zero.

This time, the end points are filled because the solutions do satisfy the inequality.

\displaystyle \left(-2\right)^{2}-4 \leq 0 \text{ and }\left(2\right)^{2}-4	\displaystyle \leq	\displaystyle 0
\displaystyle 4-4\leq 0\text{ and }4-4	\displaystyle \leq	\displaystyle 0
\displaystyle 0\leq0\text{ and }0	\displaystyle \leq	\displaystyle 0

Examples

Example 1

Solve the inequality {x^{2}}-2x-8\lt7 and graph the solution set on a number line.

Worked Solution

Create a strategy

To graph the inequality on a number line, we first want to subtract 7 from both sides. Then, we can solve the related equation for x to find the zeros.

Apply the idea

Subtracting 7 from both sides gives us {x^{2}}-2x-15\lt0.

Next, we want to solve the related equation {x^{2}}-2x-15=0 to find the zeros.

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

This gives us zeros of x=5, x=-3. To solve the inequality \left(x-5\right)\left(x+3\right)\lt0 we now want to identify the regions where this is true. Instead of drawing the entire graph, we can use a number line as the x-axis, plot the zeros, and imagine a parabola passing through those zeros. Then, we will use test points to determine which range of values makes the inequality true.

Plotting the zeros on the number line shows us that there are three regions we need to test:

The left region when x \lt -3
The middle region when -3 \lt x \lt 5
The right region when x \gt 5

A number line ranging from negative 5 to 8 in steps of 1 and with unfilled points on the negative 3 and 5 marks. A dashed vertical segment is drawn on the negative 3 and 5 marks.

We can use test points to determine which region of values would satisfy the inequality. For example, we can use x=-4 as the test point from the left region, x=1 as the test point from the middle region, and x=7 as the test point from the right region. We substitute these test points into the inequality \left(x-5\right)\left(x+3\right) \lt 0.

	\left(x-5\right)\left(x+3\right)	\text{Less than } 0?
x=-4	\left(-4-5\right)\left(-4+3\right)=9	\text{No}
x=1	\left(1-5\right)\left(1+3\right)=-16	\text{Yes}
x=7	\left(7-5\right)\left(7+3\right)=20	\text{No}

Since the inequality is less than zero, we need to look for when the answer is negative. The only negative answer comes from the test point x=1 which lies in the middle region. This tells us\left(x-5\right)\left(x+3\right) \lt 0 when -3 \lt x \lt 5. We can now graph the solution on a number line:

Reflect and check

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 \lt 5, x \lt -3 .

Using the graph to check, we see that the graph is less than zero in between the intercepts.

-4

-3

-2

-1

-18

-16

-14

-12

-10

-8

-6

-4

-2

Example 2

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

x^{2} \lt 9

Worked Solution

Create a strategy

The corresponding function is y=x^{2}.

Apply the idea

The graph of y=x^{2} is:

-4

-3

-2

-1

The solution set to the inequality x^{2} \lt 9 is equivalent to when y \lt 9 on the graph.

-4

-3

-2

-1

We can see that the graph is less than 9 between the x-values of -3 and 3. These endpoints will not be included because they do not satisfy the inequality.

The solution set for x^{2} \lt 9 is:

-3 \lt x \lt 3

Reflect and check

We could have created an equivalent inequality by subtracting 9 from both sides to get {x^{2}}-9 \lt 0. This would have translated the graph of the corresponding function down 9 units. Because one side of the inequality is 0, we can look at the x-axis rather than where the y-values are at 9.

-4

-3

-2

-1

-8

-7

-6

-5

-4

-3

-2

-1

This graph is below the x-axis when -3 \lt x \lt 3, so the solution is the same.

-x^{2}+16 \leq 0

Worked Solution

Create a strategy

The corresponding quadratic function to this inequality is y={-x^{2}}+16. This is the graph of {y=x^{2}} after it has been reflected across the x-axis and translated up 16 units.

Apply the idea

The graph of y={-x^{2}}+16 is:

-4

-3

-2

-1

-4

The solution set to the inequality -x^{2}+16 \leq 0 is equivalent to when y \leq 0 on the graph, which occurs on and below the x-axis.

-4

-3

-2

-1

-4

From the graph, we can see that the end points of the solution set for the inequality will be x=-4\text{ and }x=4

So the solution set for -x^{2}+16 \leq 0 is: x\leq -4\text{ or } x\geq 4

Reflect and check

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=5 is inside the solution set, so we can check that:

\displaystyle -x^{2}+16	\displaystyle \leq	\displaystyle 0
\displaystyle -\left(5\right)^{2}+16	\displaystyle \leq	\displaystyle 0
\displaystyle -25+16	\displaystyle \leq	\displaystyle 0
\displaystyle -9	\displaystyle \leq	\displaystyle 0

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

0 \leq \left(x-4\right)\left(x-1\right)

Worked Solution

Create a strategy

The corresponding quadratic function to this inequality is y=\left(x-4\right)\left(x-1\right). The parabola will be facing upward with x-intercepts at \left(1,0\right) and \left(4,0\right).

Apply the idea

The graph of y=\left(x-4\right)\left(x-1\right) is:

-1

-3

-2

-1

The solution set to the inequality 0\leq \left(x-4\right)\left(x-1\right) is equivalent to when 0\leq y on the graph. Another way to look at that is y\geq 0. This will be the part of the graph on or above the x-axis.

-1

-3

-2

-1

From the graph, we can see that the end points of the solution set for the inequality will be x=1 \text{ and }x=4

The solution set to 0\leq \left(x-4\right)\left(x-1\right) is : x\leq1 \text{ or }x\geq4

Reflect and check

Using the test point method, we could create a table to see when the solutions would be greater than zero or less than zero. We can use the inequality \left(x-4\right)\left(x-1\right) \geq 0 which is the same as the one we graphed above after applying the symmetric property.

	\text{Test point}	\left(x-4\right)\left(x-1\right)	\text{greater than or equal to } 0?
x\leq 1	x=0	\left(0-4\right)\left(0-1\right)=4	\text{Yes}
1\leq x\leq 4	x=2	\left(2-4\right)\left(2+1\right)=-6	\text{No}
x\geq4	x=5	\left(5-4\right)\left(5+1\right)=6	\text{Yes}

Since the inequality says the values need to be greater than zero, we look for where the values would be positive. We can see from the table this occurs when x\leq1 or when x\geq4.

Example 3

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

Worked Solution

Create a strategy

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 equal sign with the inequality symbol that would make the inequality true.

Apply the idea

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

\left(x+3\right)\left(x-1\right)=0

In the equation we created, the parabola is facing upward. We can imagine an upward facing parabola with x-intercepts at \left(-3,0\right) and \left(1,0\right). Between those points, the graph would be below the x-axis. So, the shaded region on the number line, with the end points included, represents when y\leq0.

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

Reflect and check

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. So if we had created a downward facing parabola, the inequality would have been -\left(x+3\right)\left(x-1\right) \geq 0.

Example 4

A company that produces children's toys makes a total profit, P in hundreds of dollars, given by the function P\left(x\right)=-5x^{2}+80x-315, where x is the number of toys produced in hundreds.

Write an inequality whose solution represents when the company will make a profit.

Worked Solution

Create a strategy

In order for the company to make a profit, the profit function must be positive. That is, P\left(x\right) \gt 0. We can create the inequality using the given quadratic function.

Apply the idea

-5x^{2}+80x-315 \gt 0

Reflect and check

Notice that we used \gt instead of \geq because the company has not made a profit if they have \$0.

Use the table to determine when the company would make a profit.

x	5	6	7	8	9	10	11
P\left(x\right)	-40	-15	0	5	0	-15	-40

Worked Solution

Create a strategy

We need to look for the x-values that make P\left(x\right) \gt 0. Keep in mind that the values are in hundreds.

Apply the idea

7 \lt x \lt 9

The company will make a profit if they make more than 700 toys but less than 900 toys.

Reflect and check

If the company makes less than 700 toys or more than 900 toys, they will lose money. This is likely due to the cost of the materials for the toys and the cost of producing the toys.

We can see that the company makes a maximum profit of \$500 when they make 800 toys, so this should be the number of toys that they aim to produce.

Idea summary

The solutions to a quadratic inequality are any values that make the inequality true.

When using a graph to solve a quadratic inequality with a number on one side, we look for where the y-values are equal to that number. If the inequality symbol is \lt or \leq, the solution is where the graph is below that y-value. If the inequality is \gt or \geq, the solution is where the graph is above that y-value.

Outcomes

A2.EI.2

The student will represent, solve, and interpret the solution to quadratic equations in one variable over the set of complex numbers and solve quadratic inequalities in one variable.

A2.EI.2a

Create a quadratic equation or inequality in one variable to model a contextual situation.

A2.EI.2c

Determine the solution to a quadratic inequality in one variable over the set of real numbers algebraically

A2.EI.2d

Verify possible solution(s) to quadratic equations or inequalities in one variable algebraically, graphically, and with technology to justify the reasonableness of answer(s). Explain the solution method and interpret solutions for problems given in context.

2.06 Quadratic inequalities

Ideas

Quadratic inequalities

Exploration

Examples

Example 1

Example 2

Example 3

Example 4

Outcomes

A2.EI.2

A2.EI.2a

A2.EI.2c

A2.EI.2d

What is Mathspace

About Mathspace