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Inequalities on a number line How to factor quadratic equations Solving quadratic equations Quadratic formulaHere you will learn about quadratic inequalities, including how to solve quadratic inequalities, identify solution sets using interval notation and represent solutions on a number line.

Students first learn about inequalities in 6 th grade with their work in solving linear inequalities. They expand that knowledge as they move through Algebra I and Algebra II.

**Quadratic inequalities** are mathematical statements that compare a quadratic expression to another value. Similar to linear inequalities, you can solve quadratic inequalities to find the **range of values** that makes the inequality statement true.

The strategies used to find the solutions of quadratic inequalities are the same strategies used to solve quadratic equations (factoring, using the quadratic formula, completing the square, etc).

Any quadratic equation has 0, 1 or 2 solutions only. Let’s visualize this using the quadratic equation y=x^{2}-4.

**Step-by-step guide:** Solving quadratic equations

Once you know the exact solutions to the quadratic equation, it is useful to plot these on a number line as shown above.

When solving any inequality, solutions are given in a range or interval, rather than an exact value.

For example, the inequality 4x>12 has the range of solutions x>3 as any value of x greater than 3 that is substituted into 4x will obtain a solution greater than 12. This can be shown on a number line:

You can see how by plotting the values for x on the number line, you divide the number line into two sets of numbers. Those that are greater than 3, and those that are less than or equal to 3. With a quadratic inequality, the number line can be divided into three sets of numbers.

Below are three alternative ways to determine the correct range of solutions for a quadratic inequality. Each refer to the example of y=x^{2}-4 outlined above.

The three alternatives are:

- Visualizing solution sets using graphs
- Visualizing solution sets using the coefficient of x^2
- Visualizing solution sets using substitution

Use this quiz to check your grade 1 to 7 students’ understanding of inequalities. 10+ questions with answers covering a range of 1st – 7th grade inequalities topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 1 to 7 students’ understanding of inequalities. 10+ questions with answers covering a range of 1st – 7th grade inequalities topics to identify areas of strength and support!

DOWNLOAD FREELet’s review the example of the quadratic equation y=x^{2}-4 when y=0. The two solutions of x^{2}-4=0 are x=2 and x=- \, 2 (the points of intersection between the curve y=x^{2}-4 and the straight line y=0.

Now let’s look at the quadratic equation y=x^{2}-4 when y<0. This can be described as the quadratic inequality x^{2}-4<0.

The range of values for x that gives a solution for y<0 is between - \, 2 and 2. This is the interval (- \, 2,2). Representing this on a number line gives:

Notice that both circles are open-circles as the solutions are only less than 2 or greater than - \, 2, and not equal to.

Now let’s look at the quadratic equation y=x^{2}-4 when y\geq{0}. This can be described as the quadratic inequality x^{2}-4\geq{0}.

The range of values for x that gives a solution for y\geq{0} is less than or equal to - \, 2 and greater than or equal to 2. This is the set of solutions (- \, \infty,- \, 2]\cup[2,\infty) where is the union of the two intervals. Representing this on a number line gives:

Now the two circles are closed circles as the solutions are less than or equal to - \, 2 and greater than or equal to 2.

Recall that when the coefficient of x^2 is positive, the parabola is a “ u -shape” and when the coefficient of x^2 is negative, the parabola is an “ n -shape”.

Let’s look back at the example of the quadratic equation y=x^{2}-4 when y=0. The two solutions of x^{2}-4=0 are x=2 and x=- \, 2 (the points of intersection between the curve y=x^{2}-4 and the straight line y=0.

As the quadratic y=x^{2}-4 has a positive coefficient of x^{2}\text{:}

- If y<0, then the range of solutions will be between two values. For example, x^{2}-4<0 so (- \, 2,2) .
- If y>0, then the range of solutions will be less than one value, and greater than another. For example, x^{2}-4>0 so (- \, \infty,- \, 2)\cup(2,\infty).

If the coefficient of x^{2} is negative, then the solution sets are opposite.

As the quadratic y=- \, x^{2}+4 has a negative coefficient of x^{2}\text{:}

- If y<0, then the range of solutions will be less than one value, and greater than another. For example, - \, x^{2}-4<0 so (- \, \infty,- \, 2)\cup(2,\infty).
- If y>0, then the range of solutions will be between two values. For example, - \, x^{2}-4>0 so (- \, 2,2) .

This is the same if the quadratic inequalities are also “less than or equal to”, and “greater than or equal to”.

The intervals of the two quadratics y=x^{2}-4 and y=- \, x^{2}+4 are given below when y>0 and when y<0.

You can use substitution to determine whether the solution is between two values, or the union of two intervals. Using a number line can help to visualize the correct interval(s).

Let’s look again at the example of the quadratic equation y=x^{2}-4 when y=0. The two solutions of x^{2}-4=0 are x=2 and x=- \, 2 (the points of intersection between the curve y=x^{2}-4 and the straight line y=0.

Now let’s look at the quadratic equation y=x^{2}-4 when y<0. This is the quadratic inequality x^{2}-4<0.

First, solve the quadratic equation x^{2}-4=0.

\begin{aligned}x^{2}-4&=0 \\\\ x^{2}&=4 \\\\ x&=\pm \, \sqrt{4} \\\\ x&=- \, 2\text{ or }x=2 \end{aligned}At this stage, it is not clear whether the solution is the interval (- \, 2,2) or (- \, \infty,- \, 2]\cup[2,\infty) so use substitution to find out which values of x satisfy the inequality x^{2}-4<0.

Substitute x=3 into x^{2}-4<0 to test the validity of the interval (- \, \infty,- \, 2]\cup[2,\infty).

3^{2}-4=9-4=55\nleq{0} and so this value is not within the valid range.

Conversely, substitute x=0 into x^{2}-4<0 to test its validity.

0^{2}-4=0-4=- \, 4- \, 4\leq{0} and so this value is within the valid range.

The solution set is therefore (- \, 2,2).

How does this relate to high school math?

**High School Algebra – Creating Equations (HSA-CED.A.1)**Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions

In order to solve quadratic inequalities:

**Use a strategy to solve quadratic equations to find values of \textbf{x}.****Place interval limits on the number line.****Write the intervals formed by the values of \textbf{x}.****Select a test point from the bounded interval and substitute into the quadratic.****Determine which intervals represent the solution set to the inequality.**

Determine the solution set to the quadratic inequality, x^{2}+4x+3\leq0.

**Use a strategy to solve quadratic equations to find values of \textbf{x}.**

This quadratic factors, so using the factoring method for solving quadratics, factor the quadratic to find the values of x.

\begin{aligned}&x^{2}+4x+3=0 \\\\ &(x+3)(x+1)=0 \\\\ &x+3=0\text{ so }x=- \, 3 \\\\ &\text{And }x+1=0\text{ so }x=- \,1 \end{aligned}The two solutions to the quadratic equation x^{2}+4x+3=0 are x=- \, 3 and x=- \, 1.

2**Place interval limits on the number line.**

x=- \, 3 and x=- \, 1 are the limits to the solution set for x^{2}+4x+3\leq{0}. Since the inequality is “less than or equal to” use closed points to plot the values for x.

3**Write the intervals formed by the values of \textbf{x}. **

The three intervals are:

\begin{array}{c}(- \, \infty,- \, 3] \\\\ [- \, 3,- \, 1] \\\\ [- \, 1, \infty) \end{array}4**Select a test point from the bounded interval and substitute into the quadratic.**

Let x=- \, 2\text{:}

x^{2}+4x+3=(- \, 2)^{2}+4(- \, 2)+3=4-8+3=- \, 1As - \, 1\leq{0}, this is a true comparison so the interval is part of solution.

5**Determine which intervals represent the solution set to the inequality.**

The solution set to the quadratic inequality is [- \, 3,- \, 1].

Note: Graph with number line for reference.

Solve the quadratic inequality, x^2+7 x+10>0.

**Use a strategy to solve quadratic equations to find values of \textbf{x}. **

The quadratic can be factored so use the factoring strategy of solving quadratics to find the values of x.

\begin{aligned}&x^{2}+7x+10=0 \\\\
&(x+5)(x+2)=0 \\\\
&x+5=0\text{ so }x=- \, 5 \\\\
&\text{And }x+2=0\text{ so }x=- \,2 \end{aligned}

**Place interval limits on the number line.**

Since the inequality is “greater than”, use open points to plot the limits for x.

**Write the intervals formed by the values of \textbf{x}. **

The intervals are:

\begin{array}{c}(- \, \infty,- \, 5) \\\\
(- \, 5,- \, 2) \\\\
(- \, 2,\infty) \end{array}

**Select a test point from the bounded interval and substitute into the quadratic.**

Let x=- \, 3\text{:}

x^{2}+7x+10=(- \, 3)^{2}+7(- \, 3)+10=9-21+10=- \, 2

As - \, 2\ngeq{0}, this is a false comparison so the interval is not part of solution.

**Determine which intervals represent the solution set to the inequality.**

The solution set to the quadratic inequality is (- \, \infty,- \, 5)\cup(- \, 2,\infty).

Note: Graph with number line for reference.

Solve the quadratic inequality, 5x^{2}-27x\leq{- \, 10}.

**Use a strategy to solve quadratic equations to find values of \textbf{x}. **

Before solving the quadratic for the x values, first rearrange the quadratic so that one side has a value of 0. In this case, add 10 to both sides of the inequality.

5x^{2}-27x+10\leq{0}

The quadratic can be factored to find the values of x. So use the strategy of factoring to find the values of x.

\begin{aligned}&5x^{2}-27x+10=0 \\\\ &(5x-2)(x-5)=0 \\\\ &5x-2=0\text{ so }x= \, \cfrac{2}{5} \\\\ &\text{And }x-5=0\text{ so }x=5 \end{aligned}

**Place interval limits on the number line.**

Since the inequality is “less than or equal to” use solid points to plot the limits of x.

**Write the intervals formed by the values of \textbf{x}. **

\begin{array}{c}\left(- \, \infty, \cfrac{2}{5}\right] \\\\
\left[\cfrac{2}{5}, 5\right] \\\\
[5, \infty) \end{array}

**Select a test point from the bounded interval and substitute into the quadratic.**

Let x=1\text{:}

5x^{2}-27x+10=5(1)^{2}-27(1)+10=5-27+10=- \, 12

As - \, 12\leq{0}, this is a true comparison so the interval is part of solution.

**Determine which intervals represent the solution set to the inequality.**

The solution set to the quadratic inequality is \left[\cfrac{2}{5}, 5\right].

Note: Graph with number line for reference.

Find the solution to the quadratic inequality, 2x^{2}+4x+4>- \, 5x.

**Use a strategy to solve quadratic equations to find values of \textbf{x}. **

Before solving the quadratic for the x values, first rearrange the quadratic so that one side has a value of 0. In this case, add 5x to both sides of the inequality.

\begin{aligned}2x^{2}+4x+4&>- \, 5x \\\\
2x^{2}+9x+4&>0 \end{aligned}

The quadratic can be factored to find the values of x. So use the strategy of factoring to find the values of x.

\begin{aligned}&2x^{2}+9x+4=0 \\\\
&(2x+1)(x+4)=0 \\\\
&2x+1=0\text{ so }x=- \, \cfrac{1}{2} \\\\
&\text{And }x+4=0\text{ so }x=- \, 4 \end{aligned}

**Place interval limits on the number line.**

Use open points to plot the values of x.

**Write the intervals formed by the values of \textbf{x}. **

\begin{aligned}(- \, \infty,- \ ,4) \\\\
\left(- \, 4,- \, \cfrac{1}{2}\right) \\\\
\left(- \, \cfrac{1}{2},\infty\right) \end{aligned}

**Select a test point from the bounded interval and substitute into the quadratic.**

Let x=- \, 2\text{:}

2x^{2}+9x+4=2(- \, 2)^{2}+9(- \, 2)+4=8-18+4=- \, 6

As - \, 6\ngtr{0}, this is a false comparison so the interval is not part of solution.

**Determine which intervals represent the solution set to the inequality.**

The solution set to the quadratic inequality is (- \, \infty,- \, 4)\cup\left(- \, \cfrac{1}{2},\infty\right).

Find the solution to the quadratic inequality, 4x^2\geq- \, 28x.

**Use a strategy to solve quadratic equations to find values of \textbf{x}. **

Before solving, make sure one side of the equation is 0. In this case, add 28x to both sides.

\begin{aligned}&4x^2\geq- \, 28x \\\\ &4x^2+28x\geq{0} \end{aligned}

Now you can use one of the strategies for solving quadratics. This quadratic factors, so use that strategy.

\begin{aligned}&4x^{2}+28x=0 \\\\
&4x(x+7)=0 \\\\
&4x=0\text{ so }x=0 \\\\
&\text{And }x+7=0\text{ so }x=- \, 7 \end{aligned}

**Place interval limits on the number line.**

Use closed points to plot the numbers on the number line since the original inequality is “less than and equal to.”

**Write the intervals formed by the values of \textbf{x}. **

The intervals are:

\begin{array}{c}(- \, \infty,- \, 7] \\\\
{[- \, 7,0]} \\\\
{[0,\infty)} \end{array}

**Select a test point from the bounded interval and substitute into the quadratic.**

Let x=- \, 2\text{:}

4x^{2}+28x=4(- \, 2)^{2}+28(- \, 2)=16-56=- \, 40

As - \, 40\ngeq{0}, this is a false comparison so the interval is not part of solution.

**Determine which intervals represent the solution set to the inequality.**

Since there are two intervals that represent the solution set to this quadratic inequality, use the union symbol in between intervals.

(- \, \infty,- \, 7]\cup[- \, 1,\infty)

Note: Graph with number line for reference.

Solve the quadratic inequality, 5x^2+6x-12<0.

**Use a strategy to solve quadratic equations to find values of \textbf{x}. **

The inequality is ready to solve, but notice that it does not factor. So, you can use the quadratic formula to find the values of x.

The general form of a quadratic equation is ax^2+bx+c=0.

The quadratic formula is used to solve any quadratic equation.

The quadratic formula is x=\cfrac{- \, b\pm\sqrt{b^2-4ac}}{2a}.

In this question, a=5, \, b=6, and c=- \, 12

Substitute those values into the quadratic formula to find the values for x\text{:}

\begin{aligned}x&=\cfrac{- \, 6\pm\sqrt{6^2-4\times5\times-12}}{2\times5} \\\\
&=\cfrac{- \, 6\pm\sqrt{36-(- \, 240)}}{10} \\\\
&=\cfrac{- \, 6\pm\sqrt{276}}{10} \\\\
x&=\cfrac{- \, 6+2\sqrt{69}}{10}\text{ and }x=\cfrac{- \, 6-2\sqrt{69}}{10} \end{aligned}

At this point, calculate decimals values rounded to the nearest hundredth.

x=\cfrac{- \, 6+2\sqrt{69}}{10}=1.06

x=\cfrac{- \, 6-2\sqrt{69}}{10}=- \, 2.26

**Step-by-step guide:** Quadratic formula

**Place interval limits on the number line.**

Use solid points when plotting the values of x on the number line.

**Write the intervals formed by the values of \textbf{x}. **

The intervals are:

\begin{array}{c}(- \, \infty,- \, 2.26) \\\\
(- \, 2.26,1.06) \\\\
(1.06,\infty) \end{array}

**Select a test point from the bounded interval and substitute into the quadratic.**

Let x=0\text{:}

5x^{2}+6x-12=5(0)^{2}+6(0)-12=0+0-12=- \, 12

As - \, 12\leq{0}, this is a true comparison so the interval is part of solution.

**Determine which intervals represent the solution set to the inequality.**

The intervals that are the solution set are:

(- \, 2.26,1.06)

- Start the lesson by assessing if students can recall how to solve linear inequalities, including how to graph them on the number line. This will help guide the lesson.

- Infuse activities such as gallery walks so that students have an opportunity to showcase their work as well as have open dialogue with classmates about how to solve the problems.

- Instead of worksheets, have students practice skills by game-playing.

- Have access to the plot of each quadratic equation so that students can visualize the solution.

**Forgetting the negative value when taking the square root of a number**

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.

For example:

\sqrt{16}=4 and - \, 4.4\times{4}=16 and (- \, 4)\times(- \, 4)=16.

**Not using the original inequality symbol when testing points**

When substituting test points to check if the intervals work as the solution, substitute those values for x into the original inequality.

**Mixing up way to plot numbers in the number line**

For example, when the inequality is \leq or \geq, use solid points on the number line because the solution set includes those values of x.

When the inequality is < or >, use open points on the number line because the solution set does not include the values for x.

1. Find the solution set to the quadratic inequality, x^{2}+13x+30<0.

(- \, 10, – \, 3)

[- \, 10, – \, 3]

(- \, \infty, – \, 10)

(- \, \infty, – \, 10)\cup(- \, 3, \, \infty)

The quadratic inequality factors, so use factoring to find the limit values for x.

\begin{aligned}&x^{2}+13x+30=0 \\\\ &(x+3)(x+10)=0 \\\\ &x+3=0\text{ so }x=- \, 3 \\\\ &\text{And }x+10=0\text{ so }x=- \, 10\end{aligned}

Plot the values of x on the number line and determine the intervals and the test point you will use.

Intervals:

\begin{array}{c}(- \, \infty,- \, 10) \\\\ (- \, 10,- \, 3) \\\\ (- \, 3, \infty) \end{array}

Let x=- \, 5\text{:}

x^{2}+13x+30=(- \, 5)^{2}+13(- \, 5)+30=25-65+30=- \, 10

– \,10<0 so this is a true comparison and this interval is part of the solution.

The solution is the interval: (- \, 10, – \, 3)

Note: Graph and number line for reference.

2. Solve the quadratic inequality, x^{2}+3x-10\leq{0}.

(- \, 5, 2)

(- \, \infty, – \, 5]

[- \, 5, 2]

(- \, \infty, – \, 5]\cup[2, \infty)

The quadratic inequality factors, so use factoring to find the values for x.

\begin{aligned}&x^{2}+3x-10\leq{0} \\\\ &(x+5)(x-2)=0 \\\\ &x+5=0\text{ so }x=- \, 5 \\\\ &x-2=0\text{ so }x=2 \end{aligned}

Plot the values of x on the number line and determine the intervals and the test point you will use.

Intervals:

\begin{array}{c}(- \, \infty, – \, 5] \\\\ [- \, 5, 2] \\\\ [2, \infty) \end{array}

Let x=0\text{:}

x^{2}+3x-10=(0)^{2}+3(0)-10=0+0-10=- \, 10

– \, 10\leq{0} so this is a true comparison and this interval is part of the solution.

The solution is: [- \, 5, 2]

Note: Graph and number line for reference.

3. Find the solution to the quadratic inequality, x^2+6>7 x.

(1,6)

(- \, \infty, 1)\cup(6, \infty)

[1, 6]

(- \, \infty, 1]\cup[6, \infty)

First, make sure one side of the inequality has a value of 0. In this case, subtract 7x from both sides of the inequality.

x^2-7x+6>0

The quadratic factors, so factor it to find the values of x.

\begin{aligned}&x^{2}-7x+6=0 \\\\ &(x-6)(x-1)=0 \\\\ &x-6=0\text{ so }x=6 \\\\ &x-1=0\text{ so }x=1 \end{aligned}

Plot the values of x on the number line and determine the intervals and the test point you will use.

Intervals:

\begin{array}{c}(- \, \infty,1) \\\\ (1, 6) \\\\ (6, \infty) \end{array}

Let x=2\text{:}

x^{2}-7x+6=(2)^{2}-7(2)+6=4-14+6=- \, 6

– \, 6\ngtr{0} so this is a false comparison and this interval is not part of the solution.

The solution is:

(- \, \infty, 1)\cup(6, \infty)

Note: Graph and number line for reference.

4. Find the solution to the quadratic inequality, 2x^{2}-11x+5<0.

\left[\cfrac{1}{2}, 5\right]

\left(- \, \infty,\cfrac{1}{2}\right)\cup(5, \infty)

\left(\cfrac{1}{2}, 5\right)

\left(- \, \infty, \cfrac{1}{2}]\cup[5, \infty\right)

The quadratic factors, so find the values of x by factoring.

\begin{aligned}&2x^{2}-11x+5=0 \\\\ &(2x-1)(x-5)=0 \\\\ &2x-1=0\text{ so }x=\cfrac{1}{2} \\\\ &x-5=0\text{ so }x=5 \end{aligned}

Plot the values of x on the number line and determine the intervals and the test point you will use.

Intervals:

\begin{array}{c}\left(- \, \infty, \cfrac{1}{2}\right) \\\\ \left(\cfrac{1}{2}, 5\right) \\\\ (5, \infty) \end{array}

Let x=2\text{:}

2x^{2}-11x+5=2(2)^{2}-11(2)+5=8-22+5=- \, 9

– \, 9<0 so this is a true comparison and this interval is part of the solution.

The solution is \left(\cfrac{1}{2}, 5\right)

Note: Graph and number line for reference.

5. Find the solution to the quadratic inequality, x^2-25>0.

(- \, \infty, – \, 5) \cup(5, \infty)

(- \, \infty, – \, 5] \cup[5, \infty)

[- \, 5, 5]

(- \, 5, 5)

The quadratic is the difference of two perfect squares so it factors.

\begin{aligned}&x^{2}-25=0 \\\\ &(x+5)(x-5)=0 \\\\ &x+5=0\text{ so }x=- \, 5 \\\\ &x-5=0\text{ so }x=5 \end{aligned}

Plot the values of x on the number line and determine the intervals and the test point you will use.

Intervals:

\begin{array}{c}(- \, \infty, – \, 5) \\\\ (- \, 5, 5) \\\\ (5, \infty) \end{array}

Let x=0\text{:}

x^{2}-25=(0)^{2}-25=- \, 25

– \, 25\ngtr{0} so this is a false comparison and this interval is not part of the solution.

The solution is:

(- \, \infty, – \, 5) \cup(5, \infty)

Note: Graph and number line for reference.

6. Find the solution to the quadratic inequality, 1-x^2\leq{0}.

[- \, 1, 1]

[- \, \infty, – \, 1]\cup[1, \infty]

(- \, \infty, – \, 1)\cup(1, \infty)

(- \, 1, 1)

Factor the quadratic to find the two limiting values of x\text{:}

\begin{aligned}&1-x^{2}=0 \\\\ &(1-x)(1+x)=0 \\\\ &1-x=0\text{ so }x=1 \\\\ &1+x=0\text{ so }x=- \, 1 \end{aligned}

Plot the values of x on the number line and determine the intervals and the test point you will use.

Intervals:

\begin{array}{c}(- \, \infty, – \, 1) \\\\ (- \, 1, 1) \\\\ (1, \infty) \end{array}

Let x=0\text{:}

1-x^{2}=1-(0)^{2}=1

– \, 1\nleq{0} so this is a false comparison and this interval is not part of the solution.

The solution is:

[- \, \infty, – \, 1]\cup[1, \infty]

Note: Graph and number line for reference.

7. Solve the quadratic inequality: 3x^{2}+4x-2\leq0.

[- \, 1.72, 0.39]

[- \, 0.39, 1.72]

(- \, 1.72, 0.39)

[- \, 3, 4]

The quadratic does not factor, so use the quadratic formula to find the values for x.

x=\cfrac{- \, b\pm\sqrt{b^2-4ac}}{2a}

3x^{2}+4x-2=0

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

a=3, b=4,~c=- \, 2

Substitute these values into the quadratic formula to find the values for x\text{:}

\begin{aligned}x&=\cfrac{- \, 4\pm\sqrt{4^2-4\times3\times-2}}{2\times3} \\\\ &=\cfrac{- \, 4\pm\sqrt{16-(-24)}}{6} \\\\ &=\cfrac{- \, 4\pm\sqrt{40}}{6} \\\\ x&=\cfrac{- \, 2+\sqrt{10}}{3}\text{ and }x=\cfrac{- \, 2-\sqrt{10}}{3} \end{aligned}

At this point, calculate decimals values rounded to the nearest hundredth.

x=\cfrac{- \, 2+\sqrt{10}}{3}=0.39

x=\cfrac{- \, 2-\sqrt{10}}{3}=- \, 1.72

Plot the values of x on the number line with closed points.

Intervals:

\begin{array}{c}(- \, \infty, – \, 1.72] \\\\ [- \, 1.72, 0.39] \\\\ [0.39, \infty) \end{array}

Let x=0\text{:}

3x^{2}+4x-2=3(0)^{2}+4(0)- \, 2=0+0-2=- \, 2

– \, 2\leq{0} so this is a true comparison and this interval is part of the solution.

The solution is:

[- \, 1.72, 0.39]

Note: Graph and number line for reference.

The discriminant is helpful when determining the type of solutions you will have for a quadratic equation. So, it can be helpful when solving a quadratic inequality.

Yes, you can sketch the parabola using a table of values and vertex point to determine which points make the quadratic inequality true and not true.

If the quadratic does not factor, you can use the quadratic formula or the strategy of completing the square. Remember, you can use the quadratic formula and completing the square regardless if the quadratic factors or not.

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