Solve x^2+2x+1=5

$$x ^ { 2 } + 2 x + 1 = 5$$

Solve for x (complex solution)

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\left(\sqrt{5}+1\right)\approx -3.236067977$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+2x+1=5$$

Subtract 5 from both sides of the equation.

$$x^{2}+2x+1-5=5-5$$

Subtracting 5 from itself leaves 0.

$$x^{2}+2x+1-5=0$$

Subtract 5 from 1.

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

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

$$x=\frac{-2±\sqrt{2^{2}-4\left(-4\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

Add 4 to 16.

$$x=\frac{-2±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-2±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{2} when ± is plus. Add -2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{2}$$

Divide -2+2\sqrt{5} by 2.

$$x=\sqrt{5}-1$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -2.

$$x=\frac{-2\sqrt{5}-2}{2}$$

Divide -2-2\sqrt{5} by 2.

$$x=-\sqrt{5}-1$$

The equation is now solved.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Steps for Completing the Square

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}.

$$\left(x+1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Solve for x

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\sqrt{5}-1\approx -3.236067977$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+2x+1=5$$

Subtract 5 from both sides of the equation.

$$x^{2}+2x+1-5=5-5$$

Subtracting 5 from itself leaves 0.

$$x^{2}+2x+1-5=0$$

Subtract 5 from 1.

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

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

$$x=\frac{-2±\sqrt{2^{2}-4\left(-4\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

Add 4 to 16.

$$x=\frac{-2±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-2±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{2} when ± is plus. Add -2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{2}$$

Divide -2+2\sqrt{5} by 2.

$$x=\sqrt{5}-1$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -2.

$$x=\frac{-2\sqrt{5}-2}{2}$$

Divide -2-2\sqrt{5} by 2.

$$x=-\sqrt{5}-1$$

The equation is now solved.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Steps for Completing the Square

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}.

$$\left(x+1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$