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: 108

Answer

$x=256$

Work Step by Step

$\sqrt{x}-3\sqrt[4]{x}-4=0$ Rewrite this equation using rational exponents: $x^{1/2}-3x^{1/4}-4=0$ Let $u$ be equal to $x^{1/4}$: $u=x^{1/4}$ so $u^{2}=x^{1/2}$ Substitute $x^{1/4}$ by $u$ and $x^{1/2}$ by $u^{2}$ in the original equation: $u^{2}-3u-4=0$ Solve this equation by factoring: $(u+1)(u-4)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u+1=0$ $u=-1$ $u-4=0$ $u=4$ Substitute $u$ back to $x^{1/4}$ and solve for $x$: $u=-1$ $x^{1/4}=-1$ $x=(-1)^{4}$ $x=1$ $u=4$ $x^{1/4}=4$ $x=4^{4}$ $x=256$ The solutions found are $x=1$ and $x=256$. Check them by plugging them into the original equation: $x=1$ $\sqrt{1}-3\sqrt[4]{1}-4=0$ $1-3-4=0$ $1-7=0$ $-6\ne0$ False $x=256$ $\sqrt{256}-3\sqrt[4]{256}-4=0$ $16-3(4)-4=0$ $16-12-4=0$ $0=0$ True The final answer is $x=256$
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