Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 72


The solutions are $3b$ and $-2b$, and solution set is {$3b, -2b$}.

Work Step by Step

$x^2 - bx = 6b^2$. Solve for $x$. The coefficient of the x-term is $-b$. Half of $-b$ is $-\frac{b}{2}$, and $(-\frac{b}{2})^2$ is $\frac{b^2}{4}$. Add $\frac{b^2}{4}$ to both sides of the equation to complete the square. $x^2 - bx +\frac{b^2}{4}= 6b^2+\frac{b^2}{4}$ $x^2 - bx +\frac{b^2}{4}= \frac{25b^2}{4}$ $(x-\frac{b}{2})^2 = \frac{25b^2}{4}$ $x-\frac{b}{2}= ±\sqrt\frac{25b^2}{4}$ $x = \frac{b}{2}+\frac{5b}{2}$ or $\frac{b}{2}-\frac{5b}{2}$ $x = 3b$ or $x=-2b$ The solutions are $3b$ and $-2b$, and solution set is {$3b, -2b$}.
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