8.2 Quadratic equations: solving quadratic equations

Algebra ii for the community Page 1 / 1

This module is from Elementary Algebra</link>by Denny Burzynski and Wade Ellis, Jr. Methods of solving quadratic equations as well as the logic underlying each method are discussed. Factoring, extraction of roots, completing the square, and the quadratic formula are carefully developed. The zero-factor property of real numbers is reintroduced. The chapter also includes graphs of quadratic equations based on the standard parabola, y = x^2, and applied problems from the areas of manufacturing, population, physics, geometry, mathematics (numbers and volumes), and astronomy, which are solved using the five-step method.Objectives of this module: be able to place a quadratic equation into standard form, be familiar with the zero-factor property of real numbers.

Overview

Standard Form of A Quadratic Equation
Zero-Factor Property of Real Numbers

Standard form of a quadratic equation

In Chapter ( [link] ) we studied linear equations in one and two variables and methods for solving them. We observed that a linear equation in one variable was any equation that could be written in the form $a x + b = 0, a \neq 0,$ and a linear equation in two variables was any equation that could be written in the form $a x + b y = c,$ where $a$ and $b$ are not both 0. We now wish to study quadratic equations in one variable.

Quadratic equation

A quadratic equation is an equation of the form

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

The standard form of the quadratic equation is $a x^{2} + b x + c = 0, a \neq 0.$

For a quadratic equation in standard form $a x^{2} + b x + c = 0,$

$a$ is the coefficient of $x^{2} .$
$b$ is the coefficient of $x .$
$c$ is the constant term.

Sample set a

The following are quadratic equations.

$\begin{array}{l} 3 x^{2} + 2 x - 1 = 0. & a = 3, b = 2, c = - 1. \end{array}$

$\begin{array}{l} 5 x^{2} + 8 x = 0. & a = 5, b = 8, c = 0. \end{array}$
Notice that this equation could be written $5 x^{2} + 8 x + 0 = 0.$ Now it is clear that $c = 0.$

$\begin{array}{l} x^{2} + 7 = 0. & a = 1, b = 0, c = 7. \end{array}$
Notice that this equation could be written $x^{2} + 0 x + 7 = 0.$ Now it is clear that $b = 0.$

The following are not quadratic equations.

$\begin{array}{l} 3 x + 2 = 0. & a = 0. & This equation is linear . \end{array}$

$8 x^{2} + \frac{3}{x} - 5 = 0.$
The expression on the left side of the equal sign has a variable in the denominator and, therefore, is not a quadratic.

Practice set a

Which of the following equations are quadratic equations? Answer “yes” or “no” to each equation.

$6 x^{2} - 4 x + 9 = 0$

yes

$5 x + 8 = 0$

$4 x^{3} - 5 x^{2} + x + 6 = 8$

$4 x^{2} - 2 x + 4 = 1$

yes

$\frac{2}{x} - 5 x^{2} = 6 x + 4$

$9 x^{2} - 2 x + 6 = 4 x^{2} + 8$

yes

Zero-factor property

Our goal is to solve quadratic equations. The method for solving quadratic equations is based on the zero-factor property of real numbers. We were introduced to the zero-factor property in Section [link] . We state it again.

Zero-factor property

If two numbers

a

and

b

are multiplied together and the resulting product is 0, then at least one of the numbers must be 0. Algebraically, if

a \cdot b = 0,

then

a = 0

b = 0,

or both

a = 0

and

b = 0.

Sample set b

Use the zero-factor property to solve each equation.

If $9 x = 0, then x must be 0.$

If $- 2 x^{2} = 0,$ then $x^{2} = 0, x = 0.$

If $5 (x - 1) = 0,$ then $x - 1$ must be 0, since 5 is not zero.

$\begin{array}{l} x - 1 & = & 0 \\ x & = & 1 \end{array}$

If $x (x + 6) = 0,$ then

$\begin{array}{l} x & = & 0 & or & x + 6 & = & 0 \\ x & = & - 6 \\ x & = & 0, & - 6. \end{array}$

If $(x + 2) (x + 3) = 0,$ then

$\begin{array}{l} x + 2 & = & 0 & or & x + 3 & = & 0 \\ x & = & - 2 & x & = & - 3 \\ x & = & - 2, & - 3. \end{array}$

If $(x + 10) (4 x - 5) = 0,$ then

$\begin{array}{r} x + 10 & = & 0 & or & 4 x - 5 & = & 0 \\ x & = & - 10 & 4 x & = & 5 \\ x & = & \frac{5}{4} \\ x & = & - 10, \frac{5}{4} . \end{array}$

Practice set b

Use the zero-factor property to solve each equation.

$6 (a - 4) = 0$

$a = 4$

$(y + 6) (y - 7) = 0$

$y = - 6, 7$

$(x + 5) (3 x - 4) = 0$

$x = - 5, \frac{4}{3}$

Exercises

For the following problems, write the values of $a, b,$ and $c$ in quadratic equations.

$3 x^{2} + 4 x - 7 = 0$

$3, 4, - 7$

$7 x^{2} + 2 x + 8 = 0$

$2 y^{2} - 5 y + 5 = 0$

$2, - 5, 5$

$7 a^{2} + a - 8 = 0$

$- 3 a^{2} + 4 a - 1 = 0$

$- 3, 4, - 1$

$7 b^{2} + 3 b = 0$

$2 x^{2} + 5 x = 0$

2, 5, 0

$4 y^{2} + 9 = 0$

$8 a^{2} - 2 a = 0$

$8, - 2, 0$

$6 x^{2} = 0$

$4 y^{2} = 0$

4, 0, 0

$5 x^{2} - 3 x + 9 = 4 x^{2}$

$7 x^{2} + 2 x + 1 = 6 x^{2} + x - 9$

1, 1, 10

$- 3 x^{2} + 4 x - 1 = - 4 x^{2} - 4 x + 12$

$5 x - 7 = - 3 x^{2}$

$3, 5, - 7$

$3 x - 7 = - 2 x^{2} + 5 x$

$0 = x^{2} + 6 x - 1$

$1, 6, - 1$

$9 = x^{2}$

$x^{2} = 9$

$1, 0, - 9$

$0 = - x^{2}$

For the following problems, use the zero-factor property to solve the equations.

$4 x = 0$

$x = 0$

$16 y = 0$

$9 a = 0$

$a = 0$

$4 m = 0$

$3 (k + 7) = 0$

$k = - 7$

$8 (y - 6) = 0$

$- 5 (x + 4) = 0$

$x = - 4$

$- 6 (n + 15) = 0$

$y (y - 1) = 0$

$y = 0, 1$

$a (a - 6) = 0$

$n (n + 4) = 0$

$n = 0, - 4$

$x (x + 8) = 0$

$9 (a - 4) = 0$

$a = 4$

$- 2 (m + 11) = 0$

$x (x + 7) = 0$

$x = - 7 or x = 0$

$n (n - 10) = 0$

$(y - 4) (y - 8) = 0$

$y = 4 or y = 8$

$(k - 1) (k - 6) = 0$

$(x + 5) (x + 4) = 0$

$x = - 4 or x = - 5$

$(y + 6) (2 y + 1) = 0$

$(x - 3) (5 x - 6) = 0$

$x = \frac{6}{5} or x = 3$

$(5 a + 1) (2 a - 3) = 0$

$(6 m + 5) (11 m - 6) = 0$

$m = - \frac{5}{6} or m = \frac{6}{11}$

$(2 m - 1) (3 m + 8) = 0$

$(4 x + 5) (2 x - 7) = 0$

$x = \frac{- 5}{4}, \frac{7}{2}$

$(3 y + 1) (2 y + 1) = 0$

$(7 a + 6) (7 a - 6) = 0$

$a = \frac{- 6}{7}, \frac{6}{7}$

$(8 x + 11) (2 x - 7) = 0$

$(5 x - 14) (3 x + 10) = 0$

$x = \frac{14}{5}, \frac{- 10}{3}$

$(3 x - 1) (3 x - 1) = 0$

$(2 y + 5) (2 y + 5) = 0$

$y = \frac{- 5}{2}$

${(7 a - 2)}^{2} = 0$

${(5 m - 6)}^{2} = 0$

$m = \frac{6}{5}$

Exercises for review

( [link] ) Factor $12 a x - 3 x + 8 a - 2$ by grouping.

( [link] ) Construct the graph of $6 x + 10 y - 60 = 0.$
An xy coordinate plane with gridlines, labeled negative five and five with increments of one units on both axes.

A graph of a line passing through two points coordinates zero, six and five, three.

( [link] ) Find the difference: $\frac{1}{x^{2} + 2 x + 1} - \frac{1}{x^{2} - 1} .$

( [link] ) Simplify $\sqrt{7} (\sqrt{2} + 2) .$

$\sqrt{14} + 2 \sqrt{7}$

( [link] ) Solve the radical equation $\sqrt{3 x + 10} = x + 4.$

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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