## College Algebra (11th Edition)

$x=\pm 2$
We are given: $\sqrt{2x+5}-\sqrt{x+2}=1$ $\sqrt{2x+5}=1+\sqrt{x+2}$ We square both sides: $(\sqrt{2x+5})^{2}=(1+\sqrt{x+2})^{2}$ $2x+5=1+(x+2)+2\sqrt{x+2}$ $2x+5=3+x+2\sqrt{x+2}$ $x+2=2\sqrt{x+2}$ We square both sides again: $(x+2)^{2}=(2\sqrt{x+2})^{2}$ $x^{2}+4x+4=4(x+2)$ Distribute 4 then put all terms on the left side of the equation: $x^{2}+4x+4=4x+8$ $x^2+4x+4-4x-8=0$ $x^{2}-4=0$ And factor: $(x+2)(x-2)=0$ $(x+2)=0$ or $(x-2)=0$ $x=\pm 2$