Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 148: 52


The solution is $x=8$

Work Step by Step

$\sqrt{2x}-x+4=0$ Leave the square root alone on the left side of the equation: $\sqrt{2x}=x-4$ Square both sides: $(\sqrt{2x})^{2}=(x-4)^{2}$ $2x=x^{2}-8x+16$ Take $2x$ to the right side and simplify: $0=x^{2}-8x+16-2x$ $0=x^{2}-10x+16$ Rearrange: $x^{2}-10x+16=0$ Solve by factoring: $(x-2)(x-8)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-2=0$ $x=2$ $x-8=0$ $x=8$ Check the solutions found by plugging them into the original equation: $x=2$ $\sqrt{2(2)}-2+4=0$ $\sqrt{4}-2+4=0$ $2-2+4=0$ $4\ne0$ False $x=8$ $\sqrt{2(8)}-8+4=0$ $\sqrt{16}-8+4=0$ $4-8+4=0$ $0=0$ True The solution is $x=8$
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