College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 178: 29


The solutions is $x=\dfrac{13}{6}+\dfrac{\sqrt{105}}{6}$

Work Step by Step

$\sqrt{3\sqrt{x+1}}=\sqrt{3x-5}$ Square both sides of the equation: $(\sqrt{3\sqrt{x+1}})^{2}=(\sqrt{3x-5})^{2}$ $3\sqrt{x+1}=3x-5$ Square both sides of the equation again: $(3\sqrt{x+1})^{2}=(3x-5)^{2}$ $9(x+1)=9x^{2}-30x+25$ $9x+9=9x^{2}-30x+25$ Take all terms to the right side: $0=9x^{2}-30x+25-9x-9$ Rearrange and simplify: $9x^{2}-30x+25-9x-9=0$ $9x^{2}-39x+16=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=9$, $b=-39$ and $c=16$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-39)\pm\sqrt{(-39)^{2}-4(9)(16)}}{2(9)}=...$ $...=\dfrac{39\pm\sqrt{1521-576}}{18}=\dfrac{39\pm\sqrt{945}}{18}=\dfrac{39\pm3\sqrt{105}}{18}=...$ $...=\dfrac{13}{6}\pm\dfrac{\sqrt{105}}{6}$ Verifying the solutions it can be seen that $x=\dfrac{13}{6}\pm\dfrac{\sqrt{105}}{6}$ is not a solution of the original equation. The solutions is $x=\dfrac{13}{6}+\dfrac{\sqrt{105}}{6}$
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