Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.11 - Solving Equations and Inequalities Graphically - 1.11 Exercises - Page 121: 22

Answer

$x= 0$ and $2.31$

Work Step by Step

Given equation- $1+\sqrt x$ = $\sqrt {1+x^{2}}$ , in the interval [-1,5] Substituting $\sqrt x$ = z and hence $x^{2}$ = $z^{4}$ $1+z$ = $\sqrt {1+z^{4}}$, Now squaring on both the sides- $1+z^{2} +2z $ = $1+z^{4}$ i.e. $z^{4}-z^{2}-2z$ = $0$ Graphing $y$ = $z^{4}-z^{2}-2z$, using graphing calculator- We get, $z= 0$ and $1.52$ Now $x$ = $z^{2}$ Therefore, $x$ = $0$ and $1.52^{2}$ i.e. $x$ = $0$ and $2.31$ We are asked to find all solutions 'x' that satisfy $-1 \leq x \leq 5$, Therefore in the interval [-1,5]- $x$ = $0$ and $2.31$
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