Solve Using the Quadratic Formula r^2=4r

$r^{2} = 4 r$

Step 1

Subtract $4 r$ from both sides of the equation.

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

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = - 4$ , and $c = 0$ into the quadratic formula and solve for $r$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 0)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$r = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 0$ .

Multiply $- 4$ by $1$ .

$r = \frac{4 \pm \sqrt{16 - 4 \cdot 0}}{2 \cdot 1}$

Multiply $- 4$ by $0$ .

$r = \frac{4 \pm \sqrt{16 + 0}}{2 \cdot 1}$

Add $16$ and $0$ .

$r = \frac{4 \pm \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$r = \frac{4 \pm \sqrt{4^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$r = \frac{4 \pm 4}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{4 \pm 4}{2}$

Simplify $\frac{4 \pm 4}{2}$ .

$r = 2 \pm 2$

Step 5

The final answer is the combination of both solutions.

$r = 4, 0$