Solve y^2+2y+1=2

$$y ^ { 2 } + 2 y + 1 = 2$$

Solve for y (complex solution)

$$y=\sqrt{2}-1\approx 0.414213562$$

$$y=-\left(\sqrt{2}+1\right)\approx -2.414213562$$

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.

$$y^{2}+2y+1=2$$

Subtract 2 from both sides of the equation.

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

Subtracting 2 from itself leaves 0.

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

Subtract 2 from 1.

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

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

$$y=\frac{-2±\sqrt{2^{2}-4\left(-1\right)}}{2}$$

Square 2.

$$y=\frac{-2±\sqrt{4-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$y=\frac{-2±\sqrt{4+4}}{2}$$

Add 4 to 4.

$$y=\frac{-2±\sqrt{8}}{2}$$

Take the square root of 8.

$$y=\frac{-2±2\sqrt{2}}{2}$$

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

$$y=\frac{2\sqrt{2}-2}{2}$$

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

$$y=\sqrt{2}-1$$

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

$$y=\frac{-2\sqrt{2}-2}{2}$$

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

$$y=-\sqrt{2}-1$$

The equation is now solved.

$$y=\sqrt{2}-1 y=-\sqrt{2}-1$$

Steps for Completing the Square

Factor y^{2}+2y+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y+1=\sqrt{2} y+1=-\sqrt{2}$$

Subtract 1 from both sides of the equation.

$$y=\sqrt{2}-1 y=-\sqrt{2}-1$$

Solve for y

$$y=\sqrt{2}-1\approx 0.414213562$$

$$y=-\sqrt{2}-1\approx -2.414213562$$

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.

$$y^{2}+2y+1=2$$

Subtract 2 from both sides of the equation.

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

Subtracting 2 from itself leaves 0.

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

Subtract 2 from 1.

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

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

$$y=\frac{-2±\sqrt{2^{2}-4\left(-1\right)}}{2}$$

Square 2.

$$y=\frac{-2±\sqrt{4-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$y=\frac{-2±\sqrt{4+4}}{2}$$

Add 4 to 4.

$$y=\frac{-2±\sqrt{8}}{2}$$

Take the square root of 8.

$$y=\frac{-2±2\sqrt{2}}{2}$$

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

$$y=\frac{2\sqrt{2}-2}{2}$$

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

$$y=\sqrt{2}-1$$

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

$$y=\frac{-2\sqrt{2}-2}{2}$$

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

$$y=-\sqrt{2}-1$$

The equation is now solved.

$$y=\sqrt{2}-1 y=-\sqrt{2}-1$$

Steps for Completing the Square

Factor y^{2}+2y+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y+1=\sqrt{2} y+1=-\sqrt{2}$$

Subtract 1 from both sides of the equation.

$$y=\sqrt{2}-1 y=-\sqrt{2}-1$$