## Intermediate Algebra (6th Edition)

$x=\{ 4,20 \}$
Squaring both sides of the given equation, $\sqrt{2x-4}-\sqrt{3x+4}=-2 ,$ results to \begin{array}{l}\require{cancel} (\sqrt{2x-4}-\sqrt{3x+4})^2=(-2)^2 \\ (\sqrt{2x-4})^2-2(\sqrt{2x-4})(\sqrt{3x+4})+(\sqrt{3x+4})^2=4 \\ 2x-4-2\sqrt{2x-4}\sqrt{3x+4}+3x+4=4 \\ 5x-2\sqrt{2x-4}\sqrt{3x+4}=4 \\ 5x-4=2\sqrt{2x-4}\sqrt{3x+4} .\end{array} Squaring both sides for the second time results to \begin{array}{l}\require{cancel} (5x-4)^2=(2\sqrt{2x-4}\sqrt{3x+4})^2 \\ (5x)^2-2(5x)(4)+(4)^2=4(2x-4)(3x+4) \\ 25x^2-40x+16=4(6x^2-4x-16) \\ 25x^2-40x+16=24x^2-16x-64 \\ (25x^2-24x^2)+(-40x+16x)+(16+64)=0 \\ x^2-24x+80=0 \\ (x-20)(x-4)=0 .\end{array} Equating each factor to zero (Zero Product Property) and then solving for the variable, the solutions are $x=\{ 4,20 \} .$ Upon checking, both solutions, $x=\{ 4,20 \} ,$ satisfy the original equation.