# Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5: 33

$x=25$

#### Work Step by Step

Using the properties of equality, the given equation, $4\sqrt{x}+5=x ,$ is equivalent to \begin{array}{l}\require{cancel} 4\sqrt{x}=x-5 .\end{array} Squaring both sides of the equation and then using the properties of equality, we obtain: \begin{array}{l}\require{cancel}\left( 4\sqrt{x} \right)^2=\left( x-5 \right)^2 \\\\ 16x=(x)^2+2(x)(-5)+(-5)^2 \\\\ 16x=x^2-10x+25 \\\\ 0=x^2+(-10x-16x)+25 \\\\ 0=x^2-26x+25 \\\\ (x-25)(x-1)=0 \\\\ x=\{1,25\} .\end{array} Upon checking, only $x=25$ satisfies the original equation.

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