Solve x=x^2-1

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

Solve for x

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Add 1 to both sides.

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

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.

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

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

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

Square 1.

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

Multiply -4 times -1.

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

Add 1 to 4.

$$x=\frac{-1±\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±\sqrt{5}}{-2}$$

Now solve the equation x=\frac{-1±\sqrt{5}}{-2} when ± is plus. Add -1 to \sqrt{5}.

$$x=\frac{\sqrt{5}-1}{-2}$$

Divide -1+\sqrt{5} by -2.

$$x=\frac{1-\sqrt{5}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{5}}{-2} when ± is minus. Subtract \sqrt{5} from -1.

$$x=\frac{-\sqrt{5}-1}{-2}$$

Divide -1-\sqrt{5} by -2.

$$x=\frac{\sqrt{5}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{5}}{2} x=\frac{\sqrt{5}+1}{2}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 1 by -1.

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

Divide -1 by -1.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=1+\frac{1}{4}$$

Add 1 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{5}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{5}}{2} x-\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$