## Intermediate Algebra (6th Edition)

$x=\{ 2,6 \}$
Squaring both sides of the given equation, $\sqrt{x-2}+3=\sqrt{4x+1} ,$ results to \begin{array}{l}\require{cancel} (\sqrt{x-2}+3)^2=(\sqrt{4x+1})^2 \\ (\sqrt{x-2})^2+2(\sqrt{x-2})(3)+(3)^2=4x+1 \\ x-2+6\sqrt{x-2}+9=4x+1 \\ 6\sqrt{x-2}=(4x-x)+(1+2-9) \\ 6\sqrt{x-2}=3x-6 \\ \dfrac{6\sqrt{x-2}}{3}=\dfrac{3x-6}{3} \\ 2\sqrt{x-2}=x-2 .\end{array} Squaring both sides for the second time results to \begin{array}{l}\require{cancel} (2\sqrt{x-2})^2=(x-2)^2 \\ 4(x-2)=(x)^2-2(x)(2)+(2)^2 \\ 4x-8=x^2-4x+4 \\ 0=x^2+(-4x-4x)+(4+8) \\ x^2-8x+12=0 \\ (x-6)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property) and then solving for the variable, the solutions are $x=\{ 2,6 \} .$ Upon checking, both solutions, $x=\{ 2,6 \} ,$ satisfy the original equation.