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 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set - Page 552: 150

Answer

$\{0,5\}$.

Work Step by Step

The given equation is $\Rightarrow 4x^2-16x+16=4(x+4)$ Factor out $4$ from the left hand side. $\Rightarrow 4(x^2-4x+4)=4(x+4)$ Divide the equation by $4$. $\Rightarrow \frac{4(x^2-4x+4)}{4}=\frac{4(x+4)}{4}$ Simplify. $\Rightarrow x^2-4x+4=x+4$ Add $-x-4$ to both sides. $\Rightarrow x^2-4x+4-x-4=x+4-x-4$ Add like terms. $\Rightarrow x^2-5x=0$ Factor out $x$. $\Rightarrow x(x-5)=0$ By using zero product rule set each factor equal to zero. $\Rightarrow x=0$ or $x-5=0$ Isolate $x$. $\Rightarrow x=0$ or $x=5$ The solution set is $\{0,5\}$.
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