Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.6 - Radical Equations - Exercise Set - Page 560: 40

Answer

$\{6\}$.

Work Step by Step

The given function is $f(x)=x-\sqrt{x-2}$ Replace $f(x)$ with $4$. $\Rightarrow 4=x-\sqrt{x-2}$ Add $\sqrt{x-2}-4$ to both sides. $\Rightarrow 4+\sqrt{x-2}-4=x-\sqrt{x-2}+\sqrt{x-2}-4$ Simplify. $\Rightarrow \sqrt{x-2}=x-4$ Square both sides. $\Rightarrow (\sqrt{x-2})^2=(x-4)^2$ Use the special formula $(A-B)^2=A^2-2AB+B^2$ We have $A=x$ and $B=4$ $\Rightarrow x-2=(x)^2-2(x)( 4)+(4)^2$ Simplify. $\Rightarrow x-2=x^2-8x+16$ Add $-x+2$ to both sides. $\Rightarrow x-2-x+2=x^2-8x+16-x+2$ Add like terms. $\Rightarrow 0=x^2-9x+18$ Rewrite the middle term $-9x$ as $-6x-3x$. $\Rightarrow x^2-6x-3x+18=0$ Group terms. $\Rightarrow (x^2-6x)+(-3x+18)=0$ Factor each group. $\Rightarrow x(x-6)-3(x-6)=0$ Factor out $(x-6)$. $\Rightarrow (x-6)(x-3)=0$ Set each factor equal to zero. $\Rightarrow x-6=0$ or $ x-3=0$ Isolate $x$. $\Rightarrow x=6$ or $ x=3$ Check $x=6$. $f(6)=6-\sqrt{6-2}$ $f(6)=6-\sqrt{4}$ $f(6)=6-2$ $f(6)=4$ true. Check $x=3$. $f(3)=3-\sqrt{3-2}$ $f(3)=3-\sqrt{1}$ $f(3)=3-1$ $f(3)=2$ false. The solution set is $\{6\}$.
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