Solve x^2-2x+1=2x^2-2

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

Solve for x

$$x=1$$

$$x=-3$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Add 2 to both sides.

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

Add 1 and 2 to get 3.

$$-x^{2}-2x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

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.

$$a=1 b=-3$$

Rewrite -x^{2}-2x+3 as \left(-x^{2}+x\right)+\left(-3x+3\right).

$$\left(-x^{2}+x\right)+\left(-3x+3\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(-x+1\right)+3\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(x+3\right)$$

To find equation solutions, solve -x+1=0 and x+3=0.

$$x=1 x=-3$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Add 2 to both sides.

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

Add 1 and 2 to get 3.

$$-x^{2}-2x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times -1.

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

Multiply 4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12}}{2\left(-1\right)}$$

Add 4 to 12.

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

Take the square root of 16.

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

The opposite of -2 is 2.

$$x=\frac{2±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±4}{-2}$$

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

$$x=\frac{6}{-2}$$

Divide 6 by -2.

$$x=-3$$

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

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-3 x=1$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Combine x^{2} and -2x^{2} to get -x^{2}.

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

Subtract 1 from both sides.

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

Subtract 1 from -2 to get -3.

$$-x^{2}-2x=-3$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -2 by -1.

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

Divide -3 by -1.

$$x^{2}+2x=3$$

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.

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

$$x^{2}+2x+1=4$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$