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 - Review Exercises - Page 657: 7

Answer

Given: $x^2-3x$ The coefficient of the $x$-term is $-3$. Half of $-3$ is $-\frac{3}{2}$, and $(-\frac{3}{2})^2 = \frac{9}{4}$ Thus, add $\frac{9}{4}$ to both sides to complete the square. $x^2-3x+\frac{9}{4}=\frac{9}{4}$ Since, $a^2+2ab+b^2=(a+b)^2$, the equation becomes $(x-\frac{3}{2})^2=\frac{9}{4}$ Apply the square root property to get the value of $x$. $x-\frac{3}{2}=±\sqrt{\frac{9}{4}}$ $x=\frac{3}{2}±\frac{3}{2}$ $x=\frac{3}{2}+\frac{3}{2}$ or $x=\frac{3}{2}-\frac{3}{2}$ $x=3$ or $x=0$

Work Step by Step

Given: $x^2-3x$ The coefficient of the $x$-term is $-3$. Half of $-3$ is $-\frac{3}{2}$, and $(-\frac{3}{2})^2 = \frac{9}{4}$ Thus, add $\frac{9}{4}$ to both sides to complete the square. $x^2-3x+\frac{9}{4}=\frac{9}{4}$ Since, $a^2+2ab+b^2=(a+b)^2$, the equation becomes $(x-\frac{3}{2})^2=\frac{9}{4}$ Apply the square root property to get the value of $x$. $x-\frac{3}{2}=±\sqrt{\frac{9}{4}}$ $x=\frac{3}{2}±\frac{3}{2}$ $x=\frac{3}{2}+\frac{3}{2}$ or $x=\frac{3}{2}-\frac{3}{2}$ $x=3$ or $x=0$
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