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: 48

Answer

$7$.

Work Step by Step

From the question we have $\Rightarrow \sqrt{x-3}=x-5$ Square both sides. $\Rightarrow (\sqrt{x-3})^2=(x-5)^2$ Use the special formula $(A-B)^2=A^2-2AB+B^2$ We have $A=x$ and $B=5$ $\Rightarrow x-3=(x)^2-2(x)( 5)+(5)^2$ Simplify. $\Rightarrow x-3=x^2-10x+25$ Add $-x+3$ to both sides. $\Rightarrow x-3-x+3=x^2-10x+25-x+3$ Add like terms. $\Rightarrow 0=x^2-11x+28$ Rewrite the middle term $-11x$ as $-7x-4x$. $\Rightarrow 0=x^2-7x-4x+28$ Group the terms. $\Rightarrow 0=(x^2-7x)+(-4x+28)$ Factor out each group. $\Rightarrow 0=x(x-7)-4(x-7)$ Factor out $(x-7)$. $\Rightarrow 0=(x-7)(x-4)$ Set each factor equal to zero. $\Rightarrow x-7=0$ or $x-4=0$ Isolate $x$. $\Rightarrow x=7$ or $x=4$ Check $x=7$. $\Rightarrow \sqrt{(7)-3}=7-5$ $\Rightarrow \sqrt{7-3}=2$ $\Rightarrow \sqrt{4}=2$ $\Rightarrow 2=2$ True. Check $x=4$. $\Rightarrow \sqrt{(4)-3}=4-5$ $\Rightarrow \sqrt{4-3}=-1$ $\Rightarrow \sqrt{1}=-1$ $\Rightarrow 1=-1$ false. The solution is $7$.
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