Solve x=2x^2-x-1

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

Solve for x

$$x = \frac{\sqrt{3} + 1}{2} \approx 1.366025404$$

$$x=\frac{1-\sqrt{3}}{2}\approx -0.366025404$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

Add x to both sides.

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

Combine x and x to get 2x.

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

Add 1 to both sides.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -2.

$$x=\frac{-2±\sqrt{4+8}}{2\left(-2\right)}$$

Add 4 to 8.

$$x=\frac{-2±\sqrt{12}}{2\left(-2\right)}$$

Take the square root of 12.

$$x=\frac{-2±2\sqrt{3}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-2±2\sqrt{3}}{-4}$$

Now solve the equation x=\frac{-2±2\sqrt{3}}{-4} when ± is plus. Add -2 to 2\sqrt{3}.

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

Divide -2+2\sqrt{3} by -4.

$$x=\frac{1-\sqrt{3}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{3}}{-4} when ± is minus. Subtract 2\sqrt{3} from -2.

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

Divide -2-2\sqrt{3} by -4.

$$x=\frac{\sqrt{3}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{3}}{2} x=\frac{\sqrt{3}+1}{2}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Add x to both sides.

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

Combine x and x to get 2x.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 2 by -2.

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

Divide -1 by -2.

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

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.

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

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

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

Add \frac{1}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{3}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{3}}{2} x-\frac{1}{2}=-\frac{\sqrt{3}}{2}$$

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

$$x=\frac{\sqrt{3}+1}{2} x=\frac{1-\sqrt{3}}{2}$$