Solve left(x+2right)^2=1

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

Solve for x

$$x=-1$$

$$x=-3$$

Steps Using Factoring

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

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

Subtract 1 from both sides.

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

Subtract 1 from 4 to get 3.

$$x^{2}+4x+3=0$$

To solve the equation, factor x^{2}+4x+3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=3$$

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.

$$a=1 b=3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x+1\right)\left(x+3\right)$$

To find equation solutions, solve x+1=0 and x+3=0.

$$x=-1 x=-3$$

Steps Using Factoring By Grouping

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

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

Subtract 1 from both sides.

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

Subtract 1 from 4 to get 3.

$$x^{2}+4x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 3=3$$

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.

$$a=1 b=3$$

Rewrite x^{2}+4x+3 as \left(x^{2}+x\right)+\left(3x+3\right).

$$\left(x^{2}+x\right)+\left(3x+3\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x+1\right)+3\left(x+1\right)$$

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

$$\left(x+1\right)\left(x+3\right)$$

To find equation solutions, solve x+1=0 and x+3=0.

$$x=-1 x=-3$$

Steps Using the Quadratic Formula

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

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

Subtract 1 from both sides.

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

Subtract 1 from 4 to get 3.

$$x^{2}+4x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-4±\sqrt{16-12}}{2}$$

Add 16 to -12.

$$x=\frac{-4±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-4±2}{2}$$

Now solve the equation x=\frac{-4±2}{2} when ± is plus. Add -4 to 2.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-4±2}{2} when ± is minus. Subtract 2 from -4.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=-1 x=-3$$

Steps for Completing the Square

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=1 x+2=-1$$

Subtract 2 from both sides of the equation.

$$x=-1 x=-3$$