Solve r^2-2r^2+2=r^2+4-4r

$$r ^ { 2 } - 2 r ^ { 2 } + 2 = r ^ { 2 } + 4 - 4 r$$

Solve for r

$$r=1$$

Steps Using Factoring By Grouping

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

$$-r^{2}+2=r^{2}+4-4r$$

Subtract r^{2} from both sides.

$$-r^{2}+2-r^{2}=4-4r$$

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

$$-2r^{2}+2=4-4r$$

Subtract 4 from both sides.

$$-2r^{2}+2-4=-4r$$

Subtract 4 from 2 to get -2.

$$-2r^{2}-2=-4r$$

Add 4r to both sides.

$$-2r^{2}-2+4r=0$$

Divide both sides by 2.

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

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

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

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

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

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

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

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

Factor out -r in -r^{2}+r.

$$-r\left(r-1\right)+r-1$$

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

$$\left(r-1\right)\left(-r+1\right)$$

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

$$r=1 r=1$$

Steps Using the Quadratic Formula

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

$$-r^{2}+2=r^{2}+4-4r$$

Subtract r^{2} from both sides.

$$-r^{2}+2-r^{2}=4-4r$$

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

$$-2r^{2}+2=4-4r$$

Subtract 4 from both sides.

$$-2r^{2}+2-4=-4r$$

Subtract 4 from 2 to get -2.

$$-2r^{2}-2=-4r$$

Add 4r to both sides.

$$-2r^{2}-2+4r=0$$

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.

$$-2r^{2}+4r-2=0$$

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

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

Square 4.

$$r=\frac{-4±\sqrt{16-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$r=\frac{-4±\sqrt{16+8\left(-2\right)}}{2\left(-2\right)}$$

Multiply 8 times -2.

$$r=\frac{-4±\sqrt{16-16}}{2\left(-2\right)}$$

Add 16 to -16.

$$r=\frac{-4±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

$$r=-\frac{4}{2\left(-2\right)}$$

Multiply 2 times -2.

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

Divide -4 by -4.

$$r=1$$

Steps for Completing the Square

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

$$-r^{2}+2=r^{2}+4-4r$$

Subtract r^{2} from both sides.

$$-r^{2}+2-r^{2}=4-4r$$

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

$$-2r^{2}+2=4-4r$$

Add 4r to both sides.

$$-2r^{2}+2+4r=4$$

Subtract 2 from both sides.

$$-2r^{2}+4r=4-2$$

Subtract 2 from 4 to get 2.

$$-2r^{2}+4r=2$$

Divide both sides by -2.

$$\frac{-2r^{2}+4r}{-2}=\frac{2}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$r^{2}+\frac{4}{-2}r=\frac{2}{-2}$$

Divide 4 by -2.

$$r^{2}-2r=\frac{2}{-2}$$

Divide 2 by -2.

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

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.

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

Add -1 to 1.

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

Factor r^{2}-2r+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}.

$$\left(r-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$r-1=0 r-1=0$$

Add 1 to both sides of the equation.

$$r=1 r=1$$

The equation is now solved. Solutions are the same.

$$r=1$$