Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.3 - Solving Equations by Using Quadratic Methods - Exercise Set: 2

Answer

$x=1$

Work Step by Step

$3x=\sqrt{8x+1}$ Square both sides of the equation: $(3x)^{2}=(\sqrt{8x+1})^{2}$ $9x^{2}=8x+1$ Take all terms to the left side of the equation: $9x^{2}-8x-1=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=9$, $b=-8$ and $c=-1$. Substitute: $x=\dfrac{-(-8)\pm\sqrt{(-8)^{2}-4(9)(-1)}}{2(9)}=\dfrac{8\pm\sqrt{64+36}}{18}=...$ $...=\dfrac{8\pm\sqrt{100}}{18}=\dfrac{8\pm10}{18}$ The two solutions are: $x=\dfrac{8+10}{18}=1$ $x=\dfrac{8-10}{18}=-\dfrac{1}{9}$ Verify the solutions: $3(1)=\sqrt{8(1)+1}$ $3=3$ True $3\Big(-\dfrac{1}{9}\Big)=\sqrt{8\Big(-\dfrac{1}{9}\Big)+1}$ $-\dfrac{1}{3}=\dfrac{1}{3}$ False The answer is $x=1$
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