Quadratic Equations

By the end of this section, you will be able to: solve quadratic equations by using the Zero Product Property, solve quadratic equations by factoring, and solve applications modeled by quadratic equations.

We have already solved linear equations, equations of the form $ax + by = c$ . In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. Listed below are some examples of quadratic equations:

x^2 + 5x + 6 = 0 \qquad 3y^2 + 4y = 10 \qquad 64u^2 - 81 = 0 \qquad n(n + 1) = 42

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get $n^2 + n$ . The general form of a quadratic equation is $ax^2 + bx + c = 0$ , with $a \neq 0$ .

Quadratic equation. An equation of the form

ax^2 + bx + c = 0

is called a quadratic equation, where

a

b

, and

c

are real numbers and

a \neq 0

To solve quadratic equations we need methods different than the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

Solve Quadratic Equations Using the Zero Product Property

We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Zero Product Property. If

a \cdot b = 0

, then either

a = 0

b = 0

or both.

We will now use the Zero Product Property to solve a quadratic equation.

Example. Solve: $(x + 1)(x - 4) = 0$ .

Step 1. Set each factor equal to zero. The product equals zero, so at least one factor must equal zero.

\begin{array}{lrcl} & (x + 1)(x - 4) &=& 0 \\[4pt] & x + 1 = 0 \quad\text{or}\quad x - 4 &=& 0 \end{array}

Step 2. Solve the linear equations.

x = -1 \quad\text{or}\quad x = 4

Step 3. Check. Substitute each solution separately into the original equation.

\begin{array}{rcl} (-1 + 1)(-1 - 4) &\overset{?}{=}& 0 \\ (0)(-5) &\overset{?}{=}& 0 \\ 0 &=& 0 \; \checkmark \\[6pt] (4 + 1)(4 - 4) &\overset{?}{=}& 0 \\ (5)(0) &\overset{?}{=}& 0 \\ 0 &=& 0 \; \checkmark \end{array}

The solutions are $x = -1$ and $x = 4$ .

Solve: $(x - 3)(x + 5) = 0$ . Enter the two solutions from least to greatest, separated by commas.

Solve: $(y - 6)(y + 9) = 0$ . Enter the two solutions from least to greatest, separated by commas.

We usually will do a little more work than we did in this last example to solve the linear equations that result from using the Zero Product Property.

Example. Solve: $(5n - 2)(6n - 1) = 0$ .

Use the Zero Product Property to set each factor to $0$ , then solve the equations.

\begin{array}{lrclcrcl} & 5n - 2 &=& 0 & \quad\text{or}\quad & 6n - 1 &=& 0 \\ & n &=& \tfrac{2}{5} & & n &=& \tfrac{1}{6} \end{array}

Checking both answers shows that each makes one factor equal to zero, and the product is then zero. The solutions are $n = \tfrac{2}{5}$ and $n = \tfrac{1}{6}$ .

Solve: $(3m - 2)(2m + 1) = 0$ . Enter the two solutions from least to greatest, separated by commas.

Example. Solve: $3p(10p + 7) = 0$ .

Here one of the factors is $3p$ . Use the Zero Product Property to set each factor to $0$ , then solve.

\begin{array}{lrclcrcl} & 3p &=& 0 & \quad\text{or}\quad & 10p + 7 &=& 0 \\ & p &=& 0 & & p &=& -\tfrac{7}{10} \end{array}

The solutions are $p = 0$ and $p = -\tfrac{7}{10}$ .

Solve: $2u(5u - 1) = 0$ . Enter the two solutions from least to greatest, separated by commas.

It may appear that there is only one factor in the next example. Remember, however, that $(y - 8)^2$ means $(y - 8)(y - 8)$ .

Example. Solve: $(y - 8)^2 = 0$ .

Rewrite the left side as a product, then use the Zero Product Property to set each factor to $0$ .

\begin{array}{lrcl} & (y - 8)^2 &=& 0 \\[4pt] \text{Rewrite the left side as a product.} & (y - 8)(y - 8) &=& 0 \\[4pt] \text{Set each factor to } 0. & y - 8 = 0 \quad\text{or}\quad y - 8 &=& 0 \\[4pt] \text{Solve.} & y = 8 \quad\text{or}\quad y &=& 8 \end{array}

When a solution repeats, we call it a double root. The only solution is $y = 8$ .

Solve: $(x + 1)^2 = 0$ . This has a double root — enter the single solution.

Solve Quadratic Equations by Factoring

Each of the equations we have solved in this section so far had one side in factored form. In order to use the Zero Product Property, the quadratic equation must be factored, with zero on one side. So we must be sure to start with the quadratic equation in standard form, $ax^2 + bx + c = 0$ . Then we can factor the expression on the left.

Example. Solve: $x^2 + 2x - 8 = 0$ .

Step 1. Write the quadratic equation in standard form, $ax^2 + bx + c = 0$ . The equation is already in standard form.

x^2 + 2x - 8 = 0

Step 2. Factor the quadratic expression.

(x + 4)(x - 2) = 0

Step 3. Use the Zero Product Property. Set each factor equal to zero.

x + 4 = 0 \quad\text{or}\quad x - 2 = 0

Step 4. Solve the linear equations.

x = -4 \quad\text{or}\quad x = 2

Step 5. Check. Substitute each solution separately into the original equation.

\begin{array}{rcl} (-4)^2 + 2(-4) - 8 &\overset{?}{=}& 0 \\ 16 + (-8) - 8 &\overset{?}{=}& 0 \\ 0 &=& 0 \; \checkmark \\[6pt] 2^2 + 2(2) - 8 &\overset{?}{=}& 0 \\ 4 + 4 - 8 &\overset{?}{=}& 0 \\ 0 &=& 0 \; \checkmark \end{array}

The solutions are $x = -4$ and $x = 2$ .

Solve: $x^2 - x - 12 = 0$ . Enter the two solutions from least to greatest, separated by commas.

Solve: $b^2 + 9b + 14 = 0$ . Enter the two solutions from least to greatest, separated by commas.

Solve a quadratic equation by factoring.

Write the quadratic equation in standard form, $ax^2 + bx + c = 0$ .
Factor the quadratic expression.
Use the Zero Product Property.
Solve the linear equations.
Check.

Before we factor, we must make sure the quadratic equation is in standard form.

Example. Solve: $2y^2 = 13y + 45$ .

\begin{array}{lrcl} & 2y^2 &=& 13y + 45 \\[4pt] \text{Write the quadratic equation in standard form.} & 2y^2 - 13y - 45 &=& 0 \\[4pt] \text{Factor the quadratic expression.} & (2y + 5)(y - 9) &=& 0 \\[4pt] \text{Set each factor to } 0. & 2y + 5 = 0 \quad\text{or}\quad y - 9 &=& 0 \\[4pt] \text{Solve each equation.} & y = -\tfrac{5}{2} \quad\text{or}\quad y &=& 9 \end{array}

Both answers check. The solutions are $y = -\tfrac{5}{2}$ and $y = 9$ .

Solve: $3c^2 = 10c - 8$ . Enter the two solutions from least to greatest, separated by commas.

Example. Solve: $5x^2 - 13x = 7x$ .

\begin{array}{lrcl} & 5x^2 - 13x &=& 7x \\[4pt] \text{Write the quadratic equation in standard form.} & 5x^2 - 20x &=& 0 \\[4pt] \text{Factor the left side of the equation.} & 5x(x - 4) &=& 0 \\[4pt] \text{Set each factor to } 0. & 5x = 0 \quad\text{or}\quad x - 4 &=& 0 \\[4pt] \text{Solve each equation.} & x = 0 \quad\text{or}\quad x &=& 4 \end{array}

Both answers check. The solutions are $x = 0$ and $x = 4$ .

Solve: $6a^2 + 9a = 3a$ . Enter the two solutions from least to greatest, separated by commas.

Do you recognize the special product pattern in the next example?

Example. Solve: $144q^2 = 25$ .

\begin{array}{lrcl} & 144q^2 &=& 25 \\[4pt] \text{Write the quadratic equation in standard form.} & 144q^2 - 25 &=& 0 \\[4pt] \text{Factor. It is a difference of squares.} & (12q - 5)(12q + 5) &=& 0 \\[4pt] \text{Set each factor to } 0. & 12q - 5 = 0 \quad\text{or}\quad 12q + 5 &=& 0 \\[4pt] \text{Solve each equation.} & q = \tfrac{5}{12} \quad\text{or}\quad q &=& -\tfrac{5}{12} \end{array}

The solutions are $q = \tfrac{5}{12}$ and $q = -\tfrac{5}{12}$ .

Solve: $25p^2 = 49$ . Enter the two solutions from least to greatest, separated by commas.

The left side in the next example is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

Example. Solve: $(3x - 8)(x - 1) = 3x$ .

\begin{array}{lrcl} & (3x - 8)(x - 1) &=& 3x \\[4pt] \text{Multiply the binomials.} & 3x^2 - 11x + 8 &=& 3x \\[4pt] \text{Write the quadratic equation in standard form.} & 3x^2 - 14x + 8 &=& 0 \\[4pt] \text{Factor the trinomial.} & (3x - 2)(x - 4) &=& 0 \\[4pt] \text{Set each factor to } 0. & 3x - 2 = 0 \quad\text{or}\quad x - 4 &=& 0 \\[4pt] \text{Solve each equation.} & x = \tfrac{2}{3} \quad\text{or}\quad x &=& 4 \end{array}

The solutions are $x = \tfrac{2}{3}$ and $x = 4$ .

Solve: $(2m + 1)(m + 3) = 12m$ . Enter the two solutions from least to greatest, separated by commas.

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree more than two by using the Zero Product Property, just like we solved quadratic equations.

Example. Solve: $9m^3 + 100m = 60m^2$ .

\begin{array}{lrcl} & 9m^3 + 100m &=& 60m^2 \\[4pt] \text{Bring all terms to one side so the other side is zero.} & 9m^3 - 60m^2 + 100m &=& 0 \\[4pt] \text{Factor the greatest common factor first.} & m(9m^2 - 60m + 100) &=& 0 \\[4pt] \text{Factor the trinomial.} & m(3m - 10)(3m - 10) &=& 0 \\[4pt] \text{Set each factor to } 0. & m = 0 \quad\text{or}\quad 3m - 10 &=& 0 \\[4pt] \text{Solve each equation.} & m = 0 \quad\text{or}\quad m &=& \tfrac{10}{3} \end{array}

The solutions are $m = 0$ and $m = \tfrac{10}{3}$ .

Solve: $8x^3 = 24x^2 - 18x$ . Enter the two distinct solutions from least to greatest, separated by commas.

When we factor the quadratic equation in the next example we will get three factors. However the first factor is a constant. We know that factor cannot equal $0$ .

Example. Solve: $4x^2 = 16x + 84$ .

\begin{array}{lrcl} & 4x^2 &=& 16x + 84 \\[4pt] \text{Write the quadratic equation in standard form.} & 4x^2 - 16x - 84 &=& 0 \\[4pt] \text{Factor the greatest common factor first.} & 4(x^2 - 4x - 21) &=& 0 \\[4pt] \text{Factor the trinomial.} & 4(x - 7)(x + 3) &=& 0 \\[4pt] \text{Set each factor to } 0. & x - 7 = 0 \quad\text{or}\quad x + 3 &=& 0 \\[4pt] \text{Solve each equation.} & x = 7 \quad\text{or}\quad x &=& -3 \end{array}

The factor $4$ can never equal $0$ , so it gives no solution. The solutions are $x = 7$ and $x = -3$ .

Solve: $18a^2 - 30 = -33a$ . Enter the two solutions from least to greatest, separated by commas.

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter!

Solve Applications Modeled by Quadratic Equations

The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to quadratic equations. We copy the problem-solving strategy here so we can use it for reference.

Use a problem-solving strategy to solve word problems.

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose a variable to represent that quantity.
Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then translate the English sentence into an algebra equation.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a quadratic equation.

Example. The product of two consecutive integers is $132$ . Find the integers.

Let $n =$ the first integer, so $n + 1 =$ the next consecutive integer. The first integer times the next integer is $132$ , so we translate to an equation and solve:

\begin{array}{lrcl} \text{Translate.} & n(n + 1) &=& 132 \\[4pt] \text{Multiply.} & n^2 + n &=& 132 \\[4pt] \text{Bring all the terms to one side.} & n^2 + n - 132 &=& 0 \\[4pt] \text{Factor the trinomial.} & (n - 11)(n + 12) &=& 0 \\[4pt] \text{Set each factor to } 0. & n - 11 = 0 \quad\text{or}\quad n + 12 &=& 0 \\[4pt] \text{Solve each equation.} & n = 11 \quad\text{or}\quad n &=& -12 \end{array}

There are two values for $n$ that are solutions. So there are two sets of consecutive integers that will work. If the first integer is $n = 11$ , then the next integer is $11 + 1 = 12$ . If the first integer is $n = -12$ , then the next integer is $-12 + 1 = -11$ . Both pairs check: $11 \cdot 12 = 132$ and $(-11)(-12) = 132$ . The consecutive integers are $11, 12$ and $-11, -12$ .

The product of two consecutive integers is 240. Find the two positive integers, entered from least to greatest, separated by commas.

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give $132$ . In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

Example. A rectangular garden has an area $15$ square feet. The length of the garden is two feet more than the width. Find the length and width of the garden.

Let $W =$ the width of the garden. Then $W + 2 =$ the length. The area of the rectangular garden is $15$ square feet, so we use the formula for the area of a rectangle, $A = L \cdot W$ :

\begin{array}{lrcl} \text{Substitute in the variables.} & 15 &=& (W + 2)W \\[4pt] \text{Distribute first.} & 15 &=& W^2 + 2W \\[4pt] \text{Get zero on one side.} & 0 &=& W^2 + 2W - 15 \\[4pt] \text{Factor the trinomial.} & 0 &=& (W + 5)(W - 3) \\[4pt] \text{Set each factor to } 0. & W + 5 = 0 \quad\text{or}\quad W - 3 &=& 0 \\[4pt] \text{Solve each equation.} & W = -5 \quad\text{or}\quad W &=& 3 \end{array}

Since $W$ is the width of the garden, it does not make sense for it to be negative. We eliminate that value for $W$ , so $W = 3$ . Then the length is $W + 2 = 3 + 2 = 5$ . The width of the garden is $3$ feet and the length is $5$ feet.

A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the width of the sign, in feet.

In an earlier chapter, we used the Pythagorean Theorem ( $a^2 + b^2 = c^2$ ). It gave the relation between the legs and the hypotenuse of a right triangle.

Example. Justine wants to put a deck in the corner of her backyard in the shape of a right triangle, as shown below. The hypotenuse will be $17$ feet long. The length of one side will be $7$ feet less than the length of the other side. Find the lengths of the sides of the deck.

Let $x =$ the length of one side of the deck. Then $x - 7 =$ the length of the other side. Since this is a right triangle, we use the Pythagorean Theorem:

\begin{array}{lrcl} & a^2 + b^2 &=& c^2 \\[4pt] \text{Substitute in the variables.} & x^2 + (x - 7)^2 &=& 17^2 \\[4pt] \text{Multiply.} & x^2 + x^2 - 14x + 49 &=& 289 \\[4pt] \text{Simplify.} & 2x^2 - 14x + 49 &=& 289 \\[4pt] \text{Get zero on one side.} & 2x^2 - 14x - 240 &=& 0 \\[4pt] \text{Factor the greatest common factor.} & 2(x^2 - 7x - 120) &=& 0 \\[4pt] \text{Factor the trinomial.} & 2(x - 15)(x + 8) &=& 0 \\[4pt] \text{Set each factor to } 0. & x - 15 = 0 \quad\text{or}\quad x + 8 &=& 0 \\[4pt] \text{Solve each equation.} & x = 15 \quad\text{or}\quad x &=& -8 \end{array}

Since $x$ is a side of the triangle, $x = -8$ does not make sense. So one side is $x = 15$ and the other side is $x - 7 = 15 - 7 = 8$ . Check with the Pythagorean Theorem: $15^2 + 8^2 = 225 + 64 = 289 = 17^2\ \checkmark$ . The sides of the deck are $8$ , $15$ , and $17$ feet.

A boat's sail is a right triangle. The length of one side of the sail is 7 feet more than the other side. The hypotenuse is 13 feet. Enter the lengths of the two sides (the legs) from least to greatest, separated by commas.

Key terms

quadratic equation — an equation of the form $ax^2 + bx + c = 0$ with $a \neq 0$ ; the variable is squared. Zero Product Property — if $a \cdot b = 0$ , then $a = 0$ or $b = 0$ (or both), the key fact that lets us solve a factored equation set equal to zero. double root — a solution that repeats because the same factor appears twice, as in $(y - 8)^2 = 0$ . Pythagorean Theorem — for a right triangle with legs $a$ and $b$ and hypotenuse $c$ , $a^2 + b^2 = c^2$ .

This section is adapted from Elementary Algebra 2e, Section 7.6: Quadratic Equations by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the Example step tables as prose/typeset math, recreated the right-triangle deck figure with the accessible Figure component, and described the garden sketch in prose; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

General Strategy for Factoring Polynomials Rational Expressions and Equations