Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 119

Answer

$x=20$

Work Step by Step

$\sqrt{\sqrt{x+5}+x}=5$ Square both sides: $(\sqrt{\sqrt{x+5}+x})^{2}=5^{2}$ $\sqrt{x+5}+x=25$ Take $x$ to the right side: $\sqrt{x+5}=25-x$ Again, square both sides: $(\sqrt{x+5})^{2}=(25-x)^{2}$ $x+5=625-50x+x^{2}$ Take all terms to the right side: $0=x^{2}-50x-x+625-5$ $0=x^{2}-51x+620$ Reorganize: $x^{2}-51x+620=0$ Solve this equation by factoring: $(x-20)(x-31)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-20=0$ $x=20$ $x-31=0$ $x=31$ Check your answer by plugging them into the original equation: $x=20$ $\sqrt{\sqrt{20+5}+20}=5$ $\sqrt{\sqrt{25}+20}=5$ $\sqrt{5+20}=5$ $\sqrt{25}=5$ $5=5$ True $x=31$ $\sqrt{\sqrt{31+5}+31}=5$ $\sqrt{\sqrt{36}+31}=5$ $\sqrt{6+31}=5$ $\sqrt{36}=5$ $6\ne5$ False The final answer is $x=20$
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