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\left(\sqrt{3-x}\right)^{2}=\left(x+1\right)^{2}
Square both sides of the equation.
3-x=\left(x+1\right)^{2}
Calculate \sqrt{3-x} to the power of 2 and get 3-x.
3-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}.
3-x-x^{2}=2x+1
Subtract x^{2} from both sides.
3-x-x^{2}-2x=1
Subtract 2x from both sides.
3-3x-x^{2}=1
Combine -x and -2x to get -3x.
3-3x-x^{2}-1=0
Subtract 1 from both sides.
2-3x-x^{2}=0
Subtract 1 from 3 to get 2.
-x^{2}-3x+2=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 2}}{2\left(-1\right)}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9+4\times 2}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-3\right)±\sqrt{9+8}}{2\left(-1\right)}
Multiply 4 times 2.
x=\frac{-\left(-3\right)±\sqrt{17}}{2\left(-1\right)}
Add 9 to 8.
x=\frac{3±\sqrt{17}}{2\left(-1\right)}
The opposite of -3 is 3.
x=\frac{3±\sqrt{17}}{-2}
Multiply 2 times -1.
x=\frac{\sqrt{17}+3}{-2}
Now solve the equation x=\frac{3±\sqrt{17}}{-2} when ± is plus. Add 3 to \sqrt{17}.
x=\frac{-\sqrt{17}-3}{2}
Divide 3+\sqrt{17} by -2.
x=\frac{3-\sqrt{17}}{-2}
Now solve the equation x=\frac{3±\sqrt{17}}{-2} when ± is minus. Subtract \sqrt{17} from 3.
x=\frac{\sqrt{17}-3}{2}
Divide 3-\sqrt{17} by -2.
x=\frac{-\sqrt{17}-3}{2} x=\frac{\sqrt{17}-3}{2}
The equation is now solved.
\sqrt{3-\frac{-\sqrt{17}-3}{2}}=\frac{-\sqrt{17}-3}{2}+1
Substitute \frac{-\sqrt{17}-3}{2} for x in the equation \sqrt{3-x}=x+1.
\frac{1}{2}+\frac{1}{2}\times 17^{\frac{1}{2}}=-\frac{1}{2}\times 17^{\frac{1}{2}}-\frac{1}{2}
Simplify. The value x=\frac{-\sqrt{17}-3}{2} does not satisfy the equation because the left and the right hand side have opposite signs.
\sqrt{3-\frac{\sqrt{17}-3}{2}}=\frac{\sqrt{17}-3}{2}+1
Substitute \frac{\sqrt{17}-3}{2} for x in the equation \sqrt{3-x}=x+1.
-\left(\frac{1}{2}-\frac{1}{2}\times 17^{\frac{1}{2}}\right)=\frac{1}{2}\times 17^{\frac{1}{2}}-\frac{1}{2}
Simplify. The value x=\frac{\sqrt{17}-3}{2} satisfies the equation.
x=\frac{\sqrt{17}-3}{2}
Equation \sqrt{3-x}=x+1 has a unique solution.