Solve x+1=sqrt{1-x}

$$x+1= \sqrt{ 1-x }$$

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

$$\left(x+1\right)^{2}=\left(\sqrt{1-x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1=\left(\sqrt{1-x}\right)^{2}$$

Calculate \sqrt{1-x} to the power of 2 and get 1-x.

$$x^{2}+2x+1=1-x$$

Subtract 1 from both sides.

$$x^{2}+2x+1-1=-x$$

Subtract 1 from 1 to get 0.

$$x^{2}+2x=-x$$

Add x to both sides.

$$x^{2}+2x+x=0$$

Combine 2x and x to get 3x.

$$x^{2}+3x=0$$

Factor out x.

$$x\left(x+3\right)=0$$

To find equation solutions, solve x=0 and x+3=0.

$$x=0 x=-3$$

Substitute 0 for x in the equation x+1=\sqrt{1-x}.

$$0+1=\sqrt{1-0}$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

Substitute -3 for x in the equation x+1=\sqrt{1-x}.

$$-3+1=\sqrt{1-\left(-3\right)}$$

Simplify. The value x=-3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Equation x+1=\sqrt{1-x} has a unique solution.

$$x=0$$