Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.6 Solving Quadratic Equations by Using the Quadratic Formula - 4.6 Exercises - Page 373: 37

Answer

$x= 2,x=0.5$.

Work Step by Step

The given expression is $\Rightarrow 3(4x-5)^2+20=47$ Subtract $20$ from both sides. $\Rightarrow 3(4x-5)^2+20-20=47-20$ Simplify. $\Rightarrow 3(4x-5)^2=27$ Divide both sides by $3$. $\Rightarrow \frac{3(4x-5)^2}{3}=\frac{27}{3}$ Simplify. $\Rightarrow (4x-5)^2=9$ Use the square root property. $\Rightarrow 4x-5=\pm\sqrt{9}$ Simplify. $\Rightarrow 4x-5=\pm 3$ Separate into two equaitons. $\Rightarrow 4x-5= 3$ and $4x-5=-3$ Add $5$ to each side of both equations. $\Rightarrow 4x-5+5= 3+5$ and $4x-5+5=-3+5$ Simplify. $\Rightarrow 4x= 8$ and $4x=2$ Divide each sides of both equations by $4$ $\Rightarrow \frac{4x}{4}= \frac{8}{4}$ and $\frac{4x}{4}=\frac{2}{4}$ Simplify. $\Rightarrow x= 2$ and $x=0.5$ Check both answers using the graph. The graph is shown below.
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