Answer
$\{2,6\}$.
Work Step by Step
The given expression is
$\Rightarrow \sqrt{2x-3}-\sqrt{x-2}=1$
Add $\sqrt{x-2}$ to both sides.
$\Rightarrow \sqrt{2x-3}-\sqrt{x-2}+\sqrt{x-2}=1+\sqrt{x-2}$
Simplify.
$\Rightarrow \sqrt{2x-3}=1+\sqrt{x-2}$
Square both sides.
$\Rightarrow (\sqrt{2x-3})^2=(1+\sqrt{x-2})^2$
Use the special formula $(A-B)^2=A^2-2AB+B^2$
We have $A=1$ and $B=\sqrt{x-2}$
$\Rightarrow 2x-3=(1)^2+2(1)(\sqrt{x-2})+(\sqrt{x-2})^2$
Simplify.
$\Rightarrow 2x-3=1+2\sqrt{x-2}+x-2$
$\Rightarrow 2x-3=2\sqrt{x-2}+x-1$
Add $-x+1$ to both sides.
$\Rightarrow 2x-3-x+1=2\sqrt{x-2}+x-1-x+1$
Add like terms.
$\Rightarrow x-2=2\sqrt{x-2}$
Square both sides.
$\Rightarrow (x-2)^2=(2\sqrt{x-2})^2$
Use the special formula $(A-B)^2=A^2-2AB+B^2$
We have $A=x$ and $B=2$
$\Rightarrow (x)^2-2(x)(2)+(2)^2=4(x-2)$
Simplify.
$\Rightarrow x^2-4x+4=4x-8$
Add $-4x+8$ to both sides.
$\Rightarrow x^2-4x+4-4x+8=4x-8-4x+8$
Simplify.
$\Rightarrow x^2-8x+12=0$
Rewrite the middle term $-8x$ as $-6x-2x$.
$\Rightarrow 0=x^2-6x-2x+12$
Group the terms.
$\Rightarrow 0=(x^2-6x)+(-2x+12)$
Factor each group.
$\Rightarrow 0=x(x-6)-2(x-6)$
Factor out $(x-6)$.
$\Rightarrow 0=(x-6)(x-2)$
$\Rightarrow x-6=0$ or $x-2=0$
Isolate $x$.
$\Rightarrow x=6$ or $x=2$
The solution set is $\{2,6\}$.
