Solve (x+1)^2-2=x-1

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

Solve for x

$$x=-1$$

$$x=0$$

Steps Using Factoring

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

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

Subtract 2 from 1 to get -1.

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

Subtract x from both sides.

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

Combine 2x and -x to get x.

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

Add 1 to both sides.

$$x^{2}+x-1+1=0$$

Add -1 and 1 to get 0.

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

Factor out x.

$$x\left(x+1\right)=0$$

To find equation solutions, solve x=0 and x+1=0.

$$x=0 x=-1$$

Steps Using the Quadratic Formula

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

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

Subtract 2 from 1 to get -1.

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

Subtract x from both sides.

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

Combine 2x and -x to get x.

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

Add 1 to both sides.

$$x^{2}+x-1+1=0$$

Add -1 and 1 to get 0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}}}{2}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2}$$

Now solve the equation x=\frac{-1±1}{2} when ± is plus. Add -1 to 1.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

Now solve the equation x=\frac{-1±1}{2} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{2}$$

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=0 x=-1$$

Steps for Completing the Square

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

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

Subtract 2 from 1 to get -1.

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

Subtract x from both sides.

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

Combine 2x and -x to get x.

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

Add 1 to both sides.

$$x^{2}+x-1+1=0$$

Add -1 and 1 to get 0.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{1}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{1}{2} x+\frac{1}{2}=-\frac{1}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=0 x=-1$$