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\sqrt{5-x}=x-2-1
Subtract 1 from both sides of the equation.
\sqrt{5-x}=x-3
Subtract 1 from -2 to get -3.
\left(\sqrt{5-x}\right)^{2}=\left(x-3\right)^{2}
Square both sides of the equation.
5-x=\left(x-3\right)^{2}
Calculate \sqrt{5-x} to the power of 2 and get 5-x.
5-x=x^{2}-6x+9
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.
5-x-x^{2}=-6x+9
Subtract x^{2} from both sides.
5-x-x^{2}+6x=9
Add 6x to both sides.
5+5x-x^{2}=9
Combine -x and 6x to get 5x.
5+5x-x^{2}-9=0
Subtract 9 from both sides.
-4+5x-x^{2}=0
Subtract 9 from 5 to get -4.
-x^{2}+5x-4=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=5 ab=-\left(-4\right)=4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-4. To find a and b, set up a system to be solved.
1,4 2,2
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.
1+4=5 2+2=4
Calculate the sum for each pair.
a=4 b=1
The solution is the pair that gives sum 5.
\left(-x^{2}+4x\right)+\left(x-4\right)
Rewrite -x^{2}+5x-4 as \left(-x^{2}+4x\right)+\left(x-4\right).
-x\left(x-4\right)+x-4
Factor out -x in -x^{2}+4x.
\left(x-4\right)\left(-x+1\right)
Factor out common term x-4 by using distributive property.
x=4 x=1
To find equation solutions, solve x-4=0 and -x+1=0.
\sqrt{5-4}+1=4-2
Substitute 4 for x in the equation \sqrt{5-x}+1=x-2.
2=2
Simplify. The value x=4 satisfies the equation.
\sqrt{5-1}+1=1-2
Substitute 1 for x in the equation \sqrt{5-x}+1=x-2.
3=-1
Simplify. The value x=1 does not satisfy the equation because the left and the right hand side have opposite signs.
x=4
Equation \sqrt{5-x}=x-3 has a unique solution.