Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-5 Solving Square Root and Other Radical Equations - Lesson Check - Page 394: 3

Answer

$\dfrac{1}{25}$

Work Step by Step

Subtract $7$ from both sides: $$5\sqrt{x}+7-7=8-7$$ $$5\sqrt{x}=1$$ Divide both sides by $5$: $$\frac{5\sqrt{x}}{5}=\frac{1}{5}$$ $$\sqrt{x}=\frac15$$ Square both sides: $$(\sqrt{x})^2=\left(\frac15\right)^2$$ $$x=\frac{1}{25}$$ Substitute into original problem to check for extraneous solutions: $$5\sqrt{\left(\frac{1}{25}\right)}+7=8$$ $$5\left(\frac15\right)+7=8$$ $$1+7=8$$ $$8=8$$
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