Solve (1)2x^2-1=x

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=1$$

Steps Using Factoring By Grouping

Multiply 1 and 2 to get 2.

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

Subtract x from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-1\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite 2x^{2}-x-1 as \left(2x^{2}-2x\right)+\left(x-1\right).

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

Factor out 2x in 2x^{2}-2x.

$$2x\left(x-1\right)+x-1$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(2x+1\right)$$

To find equation solutions, solve x-1=0 and 2x+1=0.

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

Steps Using the Quadratic Formula

Multiply 1 and 2 to get 2.

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

Subtract x from both sides.

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

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.

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

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

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

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2\times 2}$$

Add 1 to 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-\left(-1\right)±3}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±3}{4}$$

Now solve the equation x=\frac{1±3}{4} when ± is plus. Add 1 to 3.

$$x=\frac{4}{4}$$

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{1±3}{4} when ± is minus. Subtract 3 from 1.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply 1 and 2 to get 2.

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

Subtract x from both sides.

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

Add 1 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{3}{4} x-\frac{1}{4}=-\frac{3}{4}$$

Add \frac{1}{4} to both sides of the equation.

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