## Intermediate Algebra (6th Edition)

$x=\{ 0,4 \}$
Squaring both sides of the given equation, $\sqrt{3x+4}-1=\sqrt{2x+1} ,$ results to \begin{array}{l}\require{cancel} (\sqrt{3x+4}-1)^2=(\sqrt{2x+1})^2 \\ (\sqrt{3x+4})^2-2(\sqrt{3x+4})(1)+(1)^2=2x+1 \\ 3x+4-2\sqrt{3x+4}+1=2x+1 \\ (3x-2x)+(4+1-1)=2\sqrt{3x+4} \\ x+4=2\sqrt{3x+4} .\end{array} Squaring both sides for the second time results to \begin{array}{l}\require{cancel} (x+4)^2=(2\sqrt{3x+4})^2 \\ (x)^2+2(x)(4)+(4)^2=4(3x+4) \\ x^2+8x+16=12x+16 \\ x^2+(8x-12x)+16-16=0 \\ x^2-4x=0 \\ x(x-4)=0 .\end{array} Equating each factor to zero (Zero Product Property) and then solving for the variable, the solutions are $x=\{ 0,4 \} .$ Upon checking, both solutions, $x=\{ 0,4 \} ,$ satisfy the original equation.