Solve x+1=2x^2

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

Solve for x

$$x=1$$

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

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

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

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.

$$a+b=1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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

Subtract 2x^{2} from both sides.

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

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}.

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

Square 1.

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

Multiply -4 times -2.

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

Add 1 to 8.

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

Take the square root of 9.

$$x=\frac{-1±3}{2\left(-2\right)}$$

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.

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

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

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

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

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

Divide -4 by -4.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

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 1 by -2.

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

Divide -1 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.

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