Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 37



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given eqution, $ x^2-8x+16=0 ,$ express the equation in factored form. Then, use the Square Root Principle to solve the resulting equation. $\bf{\text{Solution Details:}}$ In the trinomial expression above, the value of $c$ is $ 16 $ and the value of $b$ is $ -8 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,16 \}, \{ 2,8 \}, \{ 4,4 \}, \\ \{ -1,-16 \}, \{ -2,-8 \}, \{ -4,-4 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -4,-4 \}.$ Hence, the factored form of the equation above is \begin{array}{l}\require{cancel} (x-4)(x-4)=0 \\\\ (x-4)^2=0 .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} x-4=\pm\sqrt{0} \\\\ x-4=0 \\\\ x=4 .\end{array}
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