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 endpoints 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:

while x^2 \geq 9 gives:

Quadratic inequality in two variables

An inequality involving two variables that contains a term of degree 2, but no term of higher degree. The solution is a set of ordered pairs represented by a region of the coordinate plane on one side of a parabola.

Example:

y \geq x^2+5x+6

We can solve quadratic inequalities in two variables by graphing the related parabola and then determining which region of the coordinate plane represents the solution.

Depending on the inequality sign, the boundary curve will be solid or dashed, and the region shaded will be above or below the boundary line.

A dashed boundary parabola that opens upward plotted in a four quadrant coordinate plane. Its minimum point or vertex is at the origin. The region above the boundary parabola is shaded.

y>x^2

A solid boundary parabola which opens upward plotted in a four quadrant coordinate plane. Its minimum point or vertex is at the origin. The region above the boundary parabola is shaded.

y \geq x^2

A dashed boundary parabola which opens upward plotted in a four quadrant coordinate plane. Its minimum point or vertex is at the origin. The region below the boundary parabola is shaded.

y<x^2

A solid boundary parabola which opens upward plotted in a four quadrant coordinate plane. The parabola opens upward with a minimum point or vertex at the origin. The region below the boundary parabola is shaded.

y \leq x^2

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 0	Factor 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 setup a table to identify whether each factor wil be positive or negative in the intervals less than, greater than, and between our two solutions of x=5 and x=3.

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

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

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 independantly. This method would give us the incorrect solution of x<5, x<-3.

Example 2

Graph the solution set for each of the following quadratic inequalities.

y \geq x^2+2

Approach

First, we want to determine what inequality sign to use. We can then draw the graph of the parabola y=x^2+2, using dashed lines if the parabola is not part of the solution itself, or a solid line if it is to be included. We then want to determine if the area above or below the parabola is to be shaded.

Solution

A solid boundary parabola representing y= x squared plus 2 plotted in a four quadrant coordinate plane. The parabola opens upward with a minimum point or vertex at ( 0, 2) . The region above the boundary parabola is shaded.

The inequality is using a greater than or equal to sign, so we will draw the parabola with a solid curve.

This also means that for every x-value, the y-value for any point in the region is greater than or equal to the y-value of the point on the parabola. So we want to shade the area above the parabola.

Reflection

It is a good idea to check our answer by substituting in a point in the region to make sure it satisfies the inequality. For example, the point \left(0,5\right) is in the region so should satisfy the inequality.

\displaystyle y	\displaystyle \geq	\displaystyle x^2+2
\displaystyle 5	\displaystyle \geq	\displaystyle 0^2+2
\displaystyle 5	\displaystyle \geq	\displaystyle 2

It does satisfy the inequality, so we have correctly drawn the graph of the inequality.

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

Approach

First, we want to identify what inequality sign is being used. We can then draw the graph of the parabola y < \left(x-5\right)\left(x+1\right), using dashed lines if the parabola is not part of the solution itself, or a solid line if it is to be included. We then want to determine if the area above or below the parabola is to be shaded.

Solution

A dashed boundary parabola representing y is less than quantity x minus 5 mutiplied by quantity x plus 1. The parabola opens upward with a minimum point or vertex at ( 2, negative 9) and passes through the points (0, negative 5), (negative 1, 0) and (5, 0) . The region below the boundary parabola is shaded.

The inequality is using a less than sign, and so we can draw the parabola with a dashed curve.

This means that for every x-value, the y-value for any point in the region is less than the y-value of the point on the parabola. So we want to shade the area below the parabola.

Reflection

It is a good idea to check our answer by substituting in a point in the region to make sure it satisfies the inequality. For example, the point \left(7, 0\right) is in the region so should satisfy the inequality.

\displaystyle y	\displaystyle <	\displaystyle \left(x-5\right)\left(x+1\right)
\displaystyle 0	\displaystyle <	\displaystyle \left(7-5\right)\left(7+1\right)
\displaystyle 0	\displaystyle <	\displaystyle 16

It does satisfy the inequality, so we have correctly drawn the graph of the inequality.

Example 3

Write the corresponding quadratic inequality for the given graph.

A solid boundary parabola plotted in a four quadrant coordinate plane. The parabola opens downwards with a maximum point or vertex at ( negative 1, 2) and pases through the points (1.5, 0), (negative 3, 0), (1, 0). The region below the boundary parabola is shaded.

Approach

First, we want to determine the equation of the boundary curve. Then we need to determine what inequality sign to use.

Solution

Since the x-values of the x-intercepts are -3 and 1, we know that the factored form will be:

y=a(x+3)(x-1)

for some value of a. We can find a by substituting in the coordinates of the vertex into the function.

To find a:

\displaystyle y	\displaystyle =	\displaystyle a(x+3)(x-1)	Factored form
\displaystyle 2	\displaystyle =	\displaystyle a(-1+3)(-1-1)	Substitute in the vertex
\displaystyle 2	\displaystyle =	\displaystyle a(2)(-2)	Simplify the factors
\displaystyle 2	\displaystyle =	\displaystyle -4a	Simplify
\displaystyle -\frac{1}{2}	\displaystyle =	\displaystyle a	Divide both sides by -4

The equation of the quadratic function in factored form:

y=-\frac{1}{2}(x+3)(x-1)

We can see that curve is solid, and that the area below the curve is shaded, so we can replace the equals sign with a less than or equal to sign.y\leq-\frac{1}{2}\left(x+3\right)\left(x-1\right)

Reflection

Alternatively, we could have found the equation in standard form: y\leq-\dfrac{x^2}{2}-x+\dfrac{3}{2}

Outcomes

M2.A.CED.A.1

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

M2.A.REI.B.2

Solve quadratic equations and inequalities in one variable.

M2.A.REI.B.2.B

Solve quadratic inequalities using the graph of the related quadratic equation.

M2.MP1

Make sense of problems and persevere in solving them.

M2.MP3

Construct viable arguments and critique the reasoning of others.

M2.MP6

Attend to precision.

M2.MP7

Look for and make use of structure.

M2.MP8

Look for and express regularity in repeated reasoning.

3.06 Quadratic inequalities

Concept summary

Worked examples

Example 1

Approach

Solution

Reflection

Example 2

Approach

Solution

Reflection

Approach

Solution

Reflection

Example 3

Approach

Solution

Reflection

Outcomes

M2.A.CED.A.1

M2.A.REI.B.2

M2.A.REI.B.2.B

M2.MP1

M2.MP3

M2.MP6

M2.MP7

M2.MP8

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