Answer
Solution set: $\{-2\}$
Work Step by Step
$\sqrt{3x+7}+\sqrt{x+2}=1$
Subtract $\sqrt{x+2}$ from both sides
$\sqrt{3x+7}=1-\sqrt{x+2}$
Square both sides
$(\sqrt{3x+7})^{2}=(1-\sqrt{x+2})^{2}$...Apply $(a-b)^{2}=a^{2}-2ab+b^{2}$
$3x+7=1-2\sqrt{x+2}+(x+2)$
$2x+4=-2\sqrt{x+2}$
Square both sides
$4x^{2}+16x+16=4(x+2)$
$4x^{2}+16x+16=4x+8$
$4x^{2}+12x+8=0$
$x^{2}+3x+2=0$
$(x+1)(x+2)=0$
Testing $x=-1,$
$\sqrt{3(-1)+7}+\sqrt{(-1)+2}=3+1=4$
... not a solution.
Testing $x=-2,$
$\sqrt{3(-2)+7}+\sqrt{(-2)+2}=0+1=1$
... $-2$ is a solution.
Solution set: $\{-2\}$
