Solve x^2=2x+1

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

Solve for x

$$x=\sqrt{2}+1\approx 2.414213562$$

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

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

Subtract 1 from both sides.

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

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}.

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

Square -2.

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

Multiply -4 times -1.

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

Add 4 to 4.

$$x=\frac{-\left(-2\right)±\sqrt{8}}{2}$$

Take the square root of 8.

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

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{2}}{2}$$

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

$$x=\frac{2\sqrt{2}+2}{2}$$

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

$$x=\sqrt{2}+1$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 2x from both sides.

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

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.

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

Add 1 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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