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Solve Quadratic Inequalities

Solve quadratic inequalities graphically

Solve Quadratic Inequalities Graphically

A quadratic equation is in standard form when written as ax² + bx + c = 0. If we replace the equal sign with an inequality sign, we have a quadratic inequality in standard form.

The graph of a quadratic function f(x) = ax² + bx + c = 0 is a parabola. When we ask when is ax² + bx + c < 0, we are asking when is f(x) < 0. We want to know when the parabola is below the x-axis.

When we ask when is ax² + bx + c > 0, we are asking when is f(x) > 0. We want to know when the parabola is above the x-axis.

How to Solve a Quadratic Inequality Graphically

Try it.

Solve \({x}^{2}-6x+8<0\) graphically. Write the solution in interval notation.

Solution

We list the steps to take to solve a quadratic inequality graphically.

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the x-axis but note how the position of the parabola affects the solution.

Example

Try it.

Solve \(\text{-}{x}^{2}-8x-12\le 0\) graphically. Write the solution in interval notation.

Solution

The quadratic inequality in standard form.		\(-{x}^{2}-8x-12\le 0\)
Graph the function \(f(x)=\text{-}{x}^{2}-8x-12\).		The parabola opens downward.
Find the line of symmetry.		\(\ x=-\frac{b}{2a}\) \(\ x=-\frac{-8}{2(-1)}\) \(\ x=-4\)
Find the vertex.		\(\ f(x)=\text{-}{x}^{2}-8x-12\) \(f(-4)=\text{-}{(-4)}^{2}-8(-4)-12\) \(f(-4)=-16+32-12\) \(f(-4)=4\) Vertex \((-4,4)\)
Find the x-intercepts. Let \(f(x)=0\).		\(\ f(x)=\text{-}{x}^{2}-8x-12\) \(\ 0=\text{-}{x}^{2}-8x-12\)
Factor. Use the Zero Product Property.		\(\ 0=-1(x+6)(x+2)\) \(\ x=-6\ x=-2\)
Graph the parabola.		x-intercepts \((-6,0),(-2,0)\)
Determine the solution from the graph. We include the x-intercepts as the inequality is “less than or equal to.”		\((\text{-}\infty ,\ \text{-}6]\cup [\text{-}2,\ \infty )\)

Solve Quadratic Inequalities Algebraically

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the zero partition numbers for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the zero partition numbers to divide the number line into intervals and then determine whether the quadratic expression will be positive or negative in the interval. We then determine the solution for the inequality.

How To Solve Quadratic Inequalities Algebraically

Try it.

Solve \({x}^{2}-x-12\ge 0\) algebraically. Write the solution in interval notation.

Solution

In this example, since the expression \({x}^{2}-x-12\) factors nicely, we can also find the sign in each interval much like we did when we solved rational inequalities. We find the sign of each of the factors, and then the sign of the product. Our number line would like this:

The result is the same as we found using the other method.

We summarize the steps here.

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation ax² + bx + c = 0. These two solutions then gave us either the two x-intercepts for the graph or the two zero partition numbers to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Solve a Quadratic Inequality Graphically
1. Write the quadratic inequality in standard form.
2. Graph the function \(f(x)=a{x}^{2}+bx+c\) using properties or transformations.
3. Determine the solution from the graph.
How to Solve a Quadratic Inequality Algebraically
1. Write the quadratic inequality in standard form.
2. Determine the zero partition numbers -- the solutions to the related quadratic equation.
3. Use the zero partition numbers to divide the number line into intervals.
4. Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve: \(2x-3=0.\)
If you missed this problem, review .

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\(x=\frac{3}{2}\)
Solve: \(2{y}^{2}+y=15\).
If you missed this problem, review .

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\(y=-3,\ y=\frac{5}{2}\)
Solve \(\frac{1}{{x}^{2}+2x-8}>0\)
If you missed this problem, review .

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\(\left(-\infty ,-4\right)\cup \left(2,\infty \right)\)
Solve \({x}^{2}-6x+8<0\) graphically. Write the solution in interval notation.
ⓐ Solve \({x}^{2}+2x-8<0\) graphically and ⓑ write the solution in interval notation.

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ⓐ

ⓑ \((-4,2)\)
ⓐ Solve \({x}^{2}-8x+12\ge 0\) graphically and ⓑ write the solution in interval notation.

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ⓐ

ⓑ \((\text{-}\infty ,2]\cup [6,\infty )\)

Solve \(\text{-}{x}^{2}-8x-12\le 0\) graphically. Write the solution in interval notation.

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The quadratic inequality in standard form.		\(-{x}^{2}-8x-12\le 0\)
Graph the function \(f(x)=\text{-}{x}^{2}-8x-12\).		The parabola opens downward.
Find the line of symmetry.		\(\ x=-\frac{b}{2a}\) \(\ x=-\frac{-8}{2(-1)}\) \(\ x=-4\)
Find the vertex.		\(\ f(x)=\text{-}{x}^{2}-8x-12\) \(f(-4)=\text{-}{(-4)}^{2}-8(-4)-12\) \(f(-4)=-16+32-12\) \(f(-4)=4\) Vertex \((-4,4)\)
Find the x-intercepts. Let \(f(x)=0\).		\(\ f(x)=\text{-}{x}^{2}-8x-12\) \(\ 0=\text{-}{x}^{2}-8x-12\)
Factor. Use the Zero Product Property.		\(\ 0=-1(x+6)(x+2)\) \(\ x=-6\ x=-2\)
Graph the parabola.		x-intercepts \((-6,0),(-2,0)\)
Determine the solution from the graph. We include the x-intercepts as the inequality is “less than or equal to.”		\((\text{-}\infty ,\ \text{-}6]\cup [\text{-}2,\ \infty )\)

ⓐ Solve \(\text{-}{x}^{2}-6x-5>0\) graphically and ⓑ write the solution in interval notation.

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ⓐ

ⓑ \((-5,-1)\)
ⓐ Solve \(\text{-}{x}^{2}+10x-16\le 0\) graphically and ⓑ write the solution in interval notation.

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ⓐ

ⓑ \((\text{-}\infty ,2]\cup [8,\infty )\)
Solve \({x}^{2}-x-12\ge 0\) algebraically. Write the solution in interval notation.
Solve \({x}^{2}+2x-8\ge 0\) algebraically. Write the solution in interval notation.

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\((\text{-}\infty ,-4]\cup [2,\infty )\)
Solve \({x}^{2}-2x-15\le 0\) algebraically. Write the solution in interval notation.

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\([-3,5]\)

Solve \({-x}^{2}+6x-7\ge 0\) algebraically. Write the solution in interval notation.

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Write the quadratic inequality in standard form.	\(-{x}^{2}+6x-7\ge 0\)
Multiply both sides of the inequality by \(-1\). Remember to reverse the inequality sign.	\(\ {x}^{2}-6x+7\le 0\)
Determine the zero partition numbers by solving the related quadratic equation.	\(\ {x}^{2}-6x+7=0\)
Write the Quadratic Formula.	\(\ x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\)
Then substitute in the values of \(a,b,c\).	\(\ x=\frac{-(-6)\pm \sqrt{{(-6)}^{2}-4⋅1⋅(7)}}{2⋅1}\)
Simplify.	\(\ x=\frac{6\pm \sqrt{8}}{2}\)
Simplify the radical.	\(\ x=\frac{6\pm 2\sqrt{2}}{2}\)
Remove the common factor, 2.	\(\ x=\frac{2(3\pm \sqrt{2})}{2}\) \(\ x=3\pm \sqrt{2}\) \(\ x=3+\sqrt{2}\ x=3-\sqrt{2}\) \(\ x\approx 1.6\ x\approx 4.4\)
Use the zero partition numbers to divide the number line into intervals. Test numbers from each interval in the original inequality.
Determine the intervals where the inequality is correct. Write the solution in interval notation.	\(\text{-}{x}^{2}+6x-7\ge 0\) in the middle interval \([3-\sqrt{2},\ 3+\sqrt{2}]\)

Solve \(\text{-}{x}^{2}+2x+1\ge 0\) algebraically. Write the solution in interval notation.

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\([-1-\sqrt{2},-1+\sqrt{2}]\)
Solve \(\text{-}{x}^{2}+8x-14<0\) algebraically. Write the solution in interval notation.

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\((\text{-}\infty ,4-\sqrt{2})\cup (4+\sqrt{2},\infty )\)

Solve, writing any solution in interval notation:

ⓐ \({x}^{2}-3x+4>0\) ⓑ \({x}^{2}-3x+4\le 0\)

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ⓐ

Write the quadratic inequality in standard form.	\(\ {x}^{2}-3x+4>0\)
Determine the zero partition numbers by solving the related quadratic equation.	\(\ {x}^{2}-3x+4=0\)
Write the Quadratic Formula.	\(x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\)
Then substitute in the values of \(a,b,c\).	\(x=\frac{-(\text{-}3)\pm \sqrt{{(\text{-}3)}^{2}-4⋅1⋅(4)}}{2⋅1}\)
Simplify.	\(x=\frac{3\pm \sqrt{-7}}{2}\)
Simplify the radicand.	\(x=\frac{3\pm \sqrt{7}i}{2}\)
The complex solutions tell us the parabola does not intercept the x-axis. Also, the parabola opens upward. This tells us that the parabola is completely above the x-axis.	Complex solutions

We are to find the solution to \({x}^{2}-3x+4>0.\) Since for all values of \(x\) the graph is above the x-axis, all values of x make the inequality true. In interval notation we write \((\text{-}\infty ,\infty ).\)

ⓑ

Write the quadratic inequality in standard form.	\({x}^{2}-3x+4\le 0\)
Determine the zero partition numbers by solving the related quadratic equation	\({x}^{2}-3x+4=0\)

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the x-axis—no part of it is below the x-axis.

We are to find the solution to \({x}^{2}-3x+4\le 0.\) Since for all values of x the graph is never below the x-axis, no values of x make the inequality true. There is no solution to the inequality.

Solve and write any solution in interval notation:
ⓐ \(\text{-}{x}^{2}+2x-4\le 0\) ⓑ \(\text{-}{x}^{2}+2x-4\ge 0\)

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ⓐ \((\text{-}\infty ,\infty )\)
ⓑ no solution
Solve and write any solution in interval notation:
ⓐ \({x}^{2}+3x+3<0\) ⓑ \({x}^{2}+3x+3>0\)

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ⓐ no solution
ⓑ \((\text{-}\infty ,\infty )\)
\({x}^{2}+6x+5>0\)

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ⓐ

ⓑ \((\text{-}\infty ,-5)\cup (-1,\infty )\)
\({x}^{2}+4x-12<0\)
\({x}^{2}+4x+3\le 0\)

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ⓐ

ⓑ \([-3,-1]\)
\({x}^{2}-6x+8\ge 0\)
\(\text{-}{x}^{2}-3x+18\le 0\)

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ⓐ

ⓑ \((\text{-}\infty ,-6]\cup [3,\infty )\)
\(\text{-}{x}^{2}+2x+24<0\)
\(\text{-}{x}^{2}+x+12\ge 0\)

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ⓐ

ⓑ \([-3,4]\)
\(\text{-}{x}^{2}+2x+15>0\)
\({x}^{2}+3x-4\ge 0\)

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\((\text{-}\infty ,-4]\cup [1,\infty )\)
\({x}^{2}+x-6\le 0\)
\({x}^{2}-7x+10<0\)

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\((2,5)\)
\({x}^{2}-4x+3>0\)
\({x}^{2}+8x>-15\)

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\((\text{-}\infty ,-5)\cup (-3,\infty )\)
\({x}^{2}+8x<-12\)
\({x}^{2}-4x+2\le 0\)

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\([2-\sqrt{2},2+\sqrt{2}]\)
\(\text{-}{x}^{2}+8x-11<0\)
\({x}^{2}-10x>-19\)

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\((\text{-}\infty ,5-\sqrt{6})\cup (5+\sqrt{6},\infty )\)
\({x}^{2}+6x<-3\)
\(-6{x}^{2}+19x-10\ge 0\)

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\(\left[\frac{2}{3},\ \frac{5}{2}\right]\)
\(-3{x}^{2}-4x+4\le 0\)
\(-2{x}^{2}+7x+4\ge 0\)

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\([\text{-}\frac{1}{2},4]\)
\(2{x}^{2}+5x-12>0\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Quadratic Inequalities

Solve quadratic inequalities graphically
Solve quadratic inequalities algebraically
Write the quadratic inequality in standard form.
Graph the function
Determine the solution from the graph.
Write the quadratic inequality in standard form.
Determine the zero partition numbers—the solutions to the related quadratic equation.
Use the zero partition numbers to divide the number line into intervals.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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