Chapter 7 - Section 7.6 - Radical Equations and Problem Solving - Exercise Set: 44

$x=5$

Work Step by Step

Squaring both sides of the given equation, $\sqrt{2x-1}-4=-\sqrt{x-4} ,$ results to \begin{array}{l}\require{cancel} (\sqrt{2x-1}-4)^2=(-\sqrt{x-4})^2 \\ (\sqrt{2x-1})^2-2(\sqrt{2x-1})(4)+(4)^2=x-4 \\ 2x-1-8\sqrt{2x-1}+16=x-4 \\ (2x-x)+(-1+16+4)=8\sqrt{2x-1} \\ x+19=8\sqrt{2x-1} .\end{array} Squaring both sides for the second time results to \begin{array}{l}\require{cancel} (x+19)^2=(8\sqrt{2x-1})^2 \\ (x)^2+2(x)(19)+(19)^2=64(2x-1) \\ x^2+38x+361=128x-64 \\ x^2+(38x-128x)+(361+64)=0 \\ x^2-90x+425=0 \\ (x-85)(x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property) and then solving for the variable, the solutions are $x=\{ 5,85 \} .$ If $x=-85,$ the original equation becomes \begin{array}{l}\require{cancel} \sqrt{2x-1}-4=-\sqrt{x-4} \\ \sqrt{2(-85)-1}-4=-\sqrt{-85-4} \\ \sqrt{-170-1}-4=-\sqrt{-89} \\ \sqrt{-171}-4=-\sqrt{-89} .\end{array} Since the radicands are negative and the indices are even, then the radicals result to nonreal numbers. Hence, only $x=5$ is the solution.

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