Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises: 23

Answer

$x=3$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ -x^2=9-6x ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} -x^2+6x-9=0 \\\\ -1(-x^2+6x-9)=0(-1) \\\\ x^2-6x+9=0 \\\\ (x-3)(x-3)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} x-3=0 \text{ OR } x-3=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} x-3=0 \\\\ x=3 \\\\\text{ OR }\\\\ x-3=0 \\\\ x=3 .\end{array} Hence, the solution is $ x=3 .$ $\bf{\text{Supplementary Solution/s:}}$ To factor the expression, $ x^2-6x+9 ,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ In the expression above, the value of $c$ is $ 9 $ and the value of $b$ is $ -6 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,9 \}, \{ 3,3 \}, \{ -1,-9 \}, \{ -3,-3 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,-3 \}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} (x-3)(x-3) .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.