Solve (2-r)^2+(1-r)^2=r^2

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

Solve for r

$$r=1$$

$$r=5$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-r\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-r\right)^{2}.

$$4-4r+r^{2}+1-2r+r^{2}=r^{2}$$

Add 4 and 1 to get 5.

$$5-4r+r^{2}-2r+r^{2}=r^{2}$$

Combine -4r and -2r to get -6r.

$$5-6r+r^{2}+r^{2}=r^{2}$$

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

$$5-6r+2r^{2}=r^{2}$$

Subtract r^{2} from both sides.

$$5-6r+2r^{2}-r^{2}=0$$

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

$$5-6r+r^{2}=0$$

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

$$r^{2}-6r+5=0$$

To solve the equation, factor r^{2}-6r+5 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=5$$

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

$$a=-5 b=-1$$

Rewrite factored expression \left(r+a\right)\left(r+b\right) using the obtained values.

$$\left(r-5\right)\left(r-1\right)$$

To find equation solutions, solve r-5=0 and r-1=0.

$$r=5 r=1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-r\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-r\right)^{2}.

$$4-4r+r^{2}+1-2r+r^{2}=r^{2}$$

Add 4 and 1 to get 5.

$$5-4r+r^{2}-2r+r^{2}=r^{2}$$

Combine -4r and -2r to get -6r.

$$5-6r+r^{2}+r^{2}=r^{2}$$

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

$$5-6r+2r^{2}=r^{2}$$

Subtract r^{2} from both sides.

$$5-6r+2r^{2}-r^{2}=0$$

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

$$5-6r+r^{2}=0$$

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

$$r^{2}-6r+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br+5. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 5=5$$

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

$$a=-5 b=-1$$

Rewrite r^{2}-6r+5 as \left(r^{2}-5r\right)+\left(-r+5\right).

$$\left(r^{2}-5r\right)+\left(-r+5\right)$$

Factor out r in the first and -1 in the second group.

$$r\left(r-5\right)-\left(r-5\right)$$

Factor out common term r-5 by using distributive property.

$$\left(r-5\right)\left(r-1\right)$$

To find equation solutions, solve r-5=0 and r-1=0.

$$r=5 r=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-r\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-r\right)^{2}.

$$4-4r+r^{2}+1-2r+r^{2}=r^{2}$$

Add 4 and 1 to get 5.

$$5-4r+r^{2}-2r+r^{2}=r^{2}$$

Combine -4r and -2r to get -6r.

$$5-6r+r^{2}+r^{2}=r^{2}$$

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

$$5-6r+2r^{2}=r^{2}$$

Subtract r^{2} from both sides.

$$5-6r+2r^{2}-r^{2}=0$$

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

$$5-6r+r^{2}=0$$

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.

$$r^{2}-6r+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5}}{2}$$

Square -6.

$$r=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2}$$

Multiply -4 times 5.

$$r=\frac{-\left(-6\right)±\sqrt{36-20}}{2}$$

Add 36 to -20.

$$r=\frac{-\left(-6\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$r=\frac{-\left(-6\right)±4}{2}$$

The opposite of -6 is 6.

$$r=\frac{6±4}{2}$$

Now solve the equation r=\frac{6±4}{2} when ± is plus. Add 6 to 4.

$$r=\frac{10}{2}$$

Divide 10 by 2.

$$r=5$$

Now solve the equation r=\frac{6±4}{2} when ± is minus. Subtract 4 from 6.

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

Divide 2 by 2.

$$r=1$$

The equation is now solved.

$$r=5 r=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-r\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-r\right)^{2}.

$$4-4r+r^{2}+1-2r+r^{2}=r^{2}$$

Add 4 and 1 to get 5.

$$5-4r+r^{2}-2r+r^{2}=r^{2}$$

Combine -4r and -2r to get -6r.

$$5-6r+r^{2}+r^{2}=r^{2}$$

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

$$5-6r+2r^{2}=r^{2}$$

Subtract r^{2} from both sides.

$$5-6r+2r^{2}-r^{2}=0$$

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

$$5-6r+r^{2}=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-6r+r^{2}=-5$$

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.

$$r^{2}-6r=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-6r+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$r^{2}-6r+9=-5+9$$

Add -5 to 9.

$$r^{2}-6r+9=4$$

Factor r^{2}-6r+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$r-3=2 r-3=-2$$

Add 3 to both sides of the equation.

$$r=5 r=1$$