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

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

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

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

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

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

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Combine 2x and 2x to get 4x.

$$2x^{2}+4x+1+1=0$$

Add 1 and 1 to get 2.

$$2x^{2}+4x+2=0$$

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\times 1=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite x^{2}+2x+1 as \left(x^{2}+x\right)+\left(x+1\right).

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

Factor out x in x^{2}+x.

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

Factor out common term x+1 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve x+1=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}.

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

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

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Combine 2x and 2x to get 4x.

$$2x^{2}+4x+1+1=0$$

Add 1 and 1 to get 2.

$$2x^{2}+4x+2=0$$

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-4±\sqrt{16-16}}{2\times 2}$$

Add 16 to -16.

$$x=\frac{-4±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$x=-\frac{4}{2\times 2}$$

Multiply 2 times 2.

$$x=-\frac{4}{4}$$

Divide -4 by 4.

$$x=-1$$

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}.

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

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

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Combine 2x and 2x to get 4x.

$$2x^{2}+4x+1+1=0$$

Add 1 and 1 to get 2.

$$2x^{2}+4x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+4x=-2$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 4 by 2.

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

Divide -2 by 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.

$$x=-1$$