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: 107

Answer

$x=\pm2\sqrt{2}$ and $x=\pm3\sqrt{3}$

Work Step by Step

$x^{4/3}-5x^{2/3}+6=0$ Let $x^{2/3}$ be equal to $u$: $u=x^{2/3}$, so $u^{2}=x^{4/3}$: Substitute $x^{4/3}$ by $u^{2}$ and $x^{2/3}$ by $u$ in the original equation: $u^{2}-5u+6=0$ Solve this equation by factoring: $(u-2)(u-3)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u-2=0$ $u=2$ $u-3=0$ $u=3$ Substitute $u$ back to $x^{2/3}$ and solve for $x$: $u=2$ $x^{2/3}=2$ $x=2^{3/2}$ $x=\sqrt{2^{3}}=\pm2\sqrt{2}$ $u=3$ $x^{2/3}=3$ $x=3^{3/2}$ $x=\sqrt{3^{3}}=\pm3\sqrt{3}$ The solutions to this equation are $x=\pm2\sqrt{2}$ and $x=\pm3\sqrt{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.