7.6 Quadratic equations

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By the end of this section, you will be able to:

Solve quadratic equations by using the Zero Product Property
Solve quadratic equations factoring
Solve applications modeled by quadratic equations

Before you get started, take this readiness quiz.

Solve: $5 y - 3 = 0$ .
If you missed this problem, review [link] .
Solve: $10 a = 0$ .
If you missed this problem, review [link] .
Combine like terms: $12 x^{2} - 6 x + 4 x$ .
If you missed this problem, review [link] .
Factor $n^{3} - 9 n^{2} - 22 n$ completely.
If you missed this problem, review [link] .

We have already solved linear equations, equations of the form $a x + b y = c$ . 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:

\begin{matrix} x^{2} + 5 x + 6 = 0 & 3 y^{2} + 4 y = 10 & 64 u^{2} - 81 = 0 & n (n + 1) = 42 \end{matrix}

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 $a x^{2} + b x + c = 0, with a \neq 0$ .

Quadratic equation

An equation of the form $a x^{2} + b x + c = 0$ is called a quadratic equation.

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 $a = 0$ or $b = 0$ or both.

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

How to use the zero product property to solve a quadratic equation

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

Solution

This table gives the steps for solving (x + 1)(x – 4) = 0. The first step is to set each factor equal to 0. Since it is a product equal to 0, at least one factor must equal 0. x + 1 = 0 or x – 4 = 0.

The last step is to check both answers by substituting the values for x into the original equation.

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Solve: $(x - 3) (x + 5) = 0$ .

$x = 3, x = −5$

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Solve: $(y - 6) (y + 9) = 0$ .

$y = 6, y = −9$

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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.

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

Solution

This image shows the steps for solving (5 n – 2)(6 n – 1) = 0. First, use the zero factor property to set each factor equal to 0, 5 n – 2 = 0 or 6 n – 1 = 0. Then, solve the equations, n = 2/5 or n = 1/6. Finally, check the answers by substituting the two solutions back into the original equation.

	$(5 n - 2) (6 n - 1) = 0$
Use the Zero Product Property to set each factor to 0.	$5 n - 2 = 0$	$6 n - 1 = 0$
Solve the equations.	$n = \frac{2}{5}$	$n = \frac{1}{6}$
Check your answers.

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Solve: $(3 m - 2) (2 m + 1) = 0$ .

$m = \frac{2}{3}, m = - \frac{1}{2}$

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Solve: $(4 p + 3) (4 p - 3) = 0$ .

$p = - \frac{3}{4}, p = \frac{3}{4}$

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Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions.

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

Solution

This image shows the steps for solving 3 p (10 p + 7) = 0. The first step is using the zero product property to set each factor equal to 0, 3p = 0 or 10 p + 7 = 0. The next step is solving both equations, p = 0 or p = negative 7/10. Finally, check the solutions by substituting the answers into the original equation.

	$3 p (10 p + 7) = 0$
Use the Zero Product Property to set each factor to 0.	$3 p = 0$	$10 p + 7 = 0$
Solve the equations.	$p = 0$	$10 p = −7$
		$p = - \frac{7}{10}$
Check your answers.

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Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand

Syamthanda Reply

hey , can you please explain oxidation reaction & redox ?

Boitumelo Reply

hey , can you please explain oxidation reaction and redox ?

Boitumelo

for grade 12 or grade 11?

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the value of V1 and V2

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advantages of electrons in a circuit

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we're do you find electromagnetism past papers

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what a normal force

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it is the force or component of the force that the surface exert on an object incontact with it and which acts perpendicular to the surface

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what is physics?

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what is the half reaction of Potassium and chlorine

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how to calculate coefficient of static friction

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how to calculate static friction

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How to calculate a current

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How to calculate force

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a structure of a thermocouple used to measure inner temperature

Anna Reply

a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4

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what is acceleration

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a rate of change in velocity of an object whith respect to time

Khuthadzo

how can we find the moment of torque of a circular object

Kidist

Acceleration is a rate of change in velocity.

Justice

t =r×f

Khuthadzo

how to calculate tension by substitution

Precious Reply

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Leago

use fnet method. how many obects are being calculated ?

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you use Fnet equals ma , newtoms second law formula

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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