Solve 0+0=1/2(x+1)^2

$$0 + 0 = \frac { 1 } { 2 } ( x + 1 ) ^ { 2 }$$

Solve for x

$$x=-1$$

Steps Using the Quadratic Formula

Add 0 and 0 to get 0.

$$0=\frac{1}{2}\left(x+1\right)^{2}$$

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

$$0=\frac{1}{2}\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply \frac{1}{2} by x^{2}+2x+1.

$$0=\frac{1}{2}x^{2}+x+\frac{1}{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}x^{2}+x+\frac{1}{2}=0$$

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

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{2}\times \frac{1}{2}}}{2\times \frac{1}{2}}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times \frac{1}{2}\times \frac{1}{2}}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-1±\sqrt{1-2\times \frac{1}{2}}}{2\times \frac{1}{2}}$$

Multiply -2 times \frac{1}{2}.

$$x=\frac{-1±\sqrt{1-1}}{2\times \frac{1}{2}}$$

Add 1 to -1.

$$x=\frac{-1±\sqrt{0}}{2\times \frac{1}{2}}$$

Take the square root of 0.

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

Multiply 2 times \frac{1}{2}.

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

Steps for Completing the Square

Add 0 and 0 to get 0.

$$0=\frac{1}{2}\left(x+1\right)^{2}$$

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

$$0=\frac{1}{2}\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply \frac{1}{2} by x^{2}+2x+1.

$$0=\frac{1}{2}x^{2}+x+\frac{1}{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}x^{2}+x+\frac{1}{2}=0$$

Subtract \frac{1}{2} from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by 2.

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

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

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

Divide 1 by \frac{1}{2} by multiplying 1 by the reciprocal of \frac{1}{2}.

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

Divide -\frac{1}{2} by \frac{1}{2} by multiplying -\frac{1}{2} by the reciprocal of \frac{1}{2}.

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

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

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

Square 1.

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

Add -1 to 1.

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

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

$$\left(x+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+1=0 x+1=0$$

Subtract 1 from both sides of the equation.

$$x=-1 x=-1$$

The equation is now solved. Solutions are the same.

$$x=-1$$