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\left(\sqrt{11-x}\right)^{2}=\left(x+1\right)^{2}
Square both sides of the equation.
11-x=\left(x+1\right)^{2}
Calculate \sqrt{11-x} to the power of 2 and get 11-x.
11-x=x^{2}+2x+1
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.
11-x-x^{2}=2x+1
Subtract x^{2} from both sides.
11-x-x^{2}-2x=1
Subtract 2x from both sides.
11-3x-x^{2}=1
Combine -x and -2x to get -3x.
11-3x-x^{2}-1=0
Subtract 1 from both sides.
10-3x-x^{2}=0
Subtract 1 from 11 to get 10.
-x^{2}-3x+10=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-3 ab=-10=-10
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+10. To find a and b, set up a system to be solved.
1,-10 2,-5
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.
1-10=-9 2-5=-3
Calculate the sum for each pair.
a=2 b=-5
The solution is the pair that gives sum -3.
\left(-x^{2}+2x\right)+\left(-5x+10\right)
Rewrite -x^{2}-3x+10 as \left(-x^{2}+2x\right)+\left(-5x+10\right).
x\left(-x+2\right)+5\left(-x+2\right)
Factor out x in the first and 5 in the second group.
\left(-x+2\right)\left(x+5\right)
Factor out common term -x+2 by using distributive property.
x=2 x=-5
To find equation solutions, solve -x+2=0 and x+5=0.
\sqrt{11-2}=2+1
Substitute 2 for x in the equation \sqrt{11-x}=x+1.
3=3
Simplify. The value x=2 satisfies the equation.
\sqrt{11-\left(-5\right)}=-5+1
Substitute -5 for x in the equation \sqrt{11-x}=x+1.
4=-4
Simplify. The value x=-5 does not satisfy the equation because the left and the right hand side have opposite signs.
x=2
Equation \sqrt{11-x}=x+1 has a unique solution.