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\sqrt{x+7}=2+\sqrt{13-x}
Subtract -\sqrt{13-x} from both sides of the equation.
\left(\sqrt{x+7}\right)^{2}=\left(2+\sqrt{13-x}\right)^{2}
Square both sides of the equation.
x+7=\left(2+\sqrt{13-x}\right)^{2}
Calculate \sqrt{x+7} to the power of 2 and get x+7.
x+7=4+4\sqrt{13-x}+\left(\sqrt{13-x}\right)^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{13-x}\right)^{2}.
x+7=4+4\sqrt{13-x}+13-x
Calculate \sqrt{13-x} to the power of 2 and get 13-x.
x+7=17+4\sqrt{13-x}-x
Add 4 and 13 to get 17.
x+7-\left(17-x\right)=4\sqrt{13-x}
Subtract 17-x from both sides of the equation.
x+7-17+x=4\sqrt{13-x}
To find the opposite of 17-x, find the opposite of each term.
x-10+x=4\sqrt{13-x}
Subtract 17 from 7 to get -10.
2x-10=4\sqrt{13-x}
Combine x and x to get 2x.
\left(2x-10\right)^{2}=\left(4\sqrt{13-x}\right)^{2}
Square both sides of the equation.
4x^{2}-40x+100=\left(4\sqrt{13-x}\right)^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-10\right)^{2}.
4x^{2}-40x+100=4^{2}\left(\sqrt{13-x}\right)^{2}
Expand \left(4\sqrt{13-x}\right)^{2}.
4x^{2}-40x+100=16\left(\sqrt{13-x}\right)^{2}
Calculate 4 to the power of 2 and get 16.
4x^{2}-40x+100=16\left(13-x\right)
Calculate \sqrt{13-x} to the power of 2 and get 13-x.
4x^{2}-40x+100=208-16x
Use the distributive property to multiply 16 by 13-x.
4x^{2}-40x+100-208=-16x
Subtract 208 from both sides.
4x^{2}-40x-108=-16x
Subtract 208 from 100 to get -108.
4x^{2}-40x-108+16x=0
Add 16x to both sides.
4x^{2}-24x-108=0
Combine -40x and 16x to get -24x.
x^{2}-6x-27=0
Divide both sides by 4.
a+b=-6 ab=1\left(-27\right)=-27
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-27. To find a and b, set up a system to be solved.
1,-27 3,-9
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 -27.
1-27=-26 3-9=-6
Calculate the sum for each pair.
a=-9 b=3
The solution is the pair that gives sum -6.
\left(x^{2}-9x\right)+\left(3x-27\right)
Rewrite x^{2}-6x-27 as \left(x^{2}-9x\right)+\left(3x-27\right).
x\left(x-9\right)+3\left(x-9\right)
Factor out x in the first and 3 in the second group.
\left(x-9\right)\left(x+3\right)
Factor out common term x-9 by using distributive property.
x=9 x=-3
To find equation solutions, solve x-9=0 and x+3=0.
\sqrt{9+7}-\sqrt{13-9}=2
Substitute 9 for x in the equation \sqrt{x+7}-\sqrt{13-x}=2.
2=2
Simplify. The value x=9 satisfies the equation.
\sqrt{-3+7}-\sqrt{13-\left(-3\right)}=2
Substitute -3 for x in the equation \sqrt{x+7}-\sqrt{13-x}=2.
-2=2
Simplify. The value x=-3 does not satisfy the equation because the left and the right hand side have opposite signs.
\sqrt{9+7}-\sqrt{13-9}=2
Substitute 9 for x in the equation \sqrt{x+7}-\sqrt{13-x}=2.
2=2
Simplify. The value x=9 satisfies the equation.
x=9
Equation \sqrt{x+7}=\sqrt{13-x}+2 has a unique solution.