Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.6 - Radical Equations and Problem Solving - Exercise Set: 43

Answer

$x = \{ 1, \frac{1}{4}\}$

Work Step by Step

$\sqrt (5x-1) -\sqrt x + 2 = 3 $ To separate the radicals add $-2$ both sides. $\sqrt (5x-1) -\sqrt x +2 - 2 = 3 - 2 $ $\sqrt (5x-1) -\sqrt x = 1 $ Squaring on both sides. $(\sqrt (5x-1) -\sqrt x)^{2} = 1^{2} $ $(5x-1) + x - 2 \sqrt (5x-1)\sqrt x = 1$ $6x-1 - 2 \sqrt (5x-1)\sqrt x = 1$ $6x-1 -1 = 2 \sqrt (5x-1)\sqrt x $ $6x-2 = 2 \sqrt (5x-1)\sqrt x $ $2(3x-1) = 2 \sqrt (5x-1)\sqrt x $ Dividing both sides by $2$ $(3x-1) = \sqrt (5x-1)\sqrt x $ Squaring on both sides. $(3x-1)^{2} = (\sqrt (5x-1)\sqrt x )^{2}$ $9x^{2}-6x+1 = x(5x-1)$ $9x^{2}-6x+1 = 5x^{2}-x$ $9x^{2}-6x+1 - 5x^{2}+x = 0$ $4x^{2}-5x+1 = 0$ By factoring, $(4x-1)(x-1)=0$ $x = 1$ or $x= \frac{1}{4}$ $x = \{ 1, \frac{1}{4}\}$ Substituting $x$ values in the given equation, $x= 1$ satisfies the equation. So, $1$ is the only solution.
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