Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.3 - Solving Equations by Using Quadratic Methods - Exercise Set: 6

Answer

$x=9$ and $x=1$

Work Step by Step

$\sqrt{16x}=x+3$ Square both sides of the equation: $(\sqrt{16x})^{2}=(x+3)^{2}$ $16x=x^{2}+6x+9$ Take $16x$ to the right side of the equation and simplify it by combining like terms: $0=x^{2}+6x-16x+9$ $x^{2}-10x+9=0$ Solve this equation by factoring: $(x-9)(x-1)=0$ Set both factors equal to $0$ and solve each individual equation: $x-9=0$ $x=9$ $x-1=0$ $x=1$ Check the answers found by substituting them into the original equation: $x=9$ $\sqrt{16(9)}=9+3$ $12=12$ True $x=1$ $\sqrt{16(1)}=1+3$ $4=4$ True The final answers are $x=9$ and $x=1$
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