Answer
$x=27$
Work Step by Step
We are given:
$\sqrt{2x-5}=2+\sqrt{x-2}$
We square both sides:
$(\sqrt{2x-5})^{2}=(2+\sqrt{x-2})^{2}$
$(\sqrt{2x-5})(\sqrt{2x-5})=(2+\sqrt{x-2})(2+\sqrt{x-2})$
$2x-5=4+4\sqrt{x-2}+x-2$
$2x-5-2=x+4\sqrt{x-2}$
$x-7=4\sqrt{x-2}$
We square both sides again:
$(x-7)^{2}=(4\sqrt{x-2})^{2}$
$x^{2}-14x+49=16(x-2)$
And distribute:
$x^{2}-14x+49=16x-32$
$x^{2}-30x+81=0$
And factor:
$(x-3)(x-27)=0$
Use the zero-factor property by equating each factor to zero:
$(x-3)=0$ or $(x-27)=0$
$x=3$ or $x=27$
However, the solution $x=3$ does not work in the original equation:
$\sqrt{2(3)-5}=2+\sqrt{3-2}$
$\sqrt{6-5}=2+\sqrt{1}$
$\sqrt{1}=2+1$
$1=3$
Since we got a false statement, $x=3$ is not a solution. Thus the only solution is
$x=27$.