# How do you solve r= \sqrt { - 3+ 4r }?

Mar 13, 2018

$r = 1 \mathmr{and} r = 3$

#### Explanation:

$r = \sqrt{- 3 + 4 r}$

square both sides:

${r}^{2} = - 3 + 4 r$

add $3$ and subtract $4 r$, to make the RHS $0 :$

${r}^{2} - 4 r + 3 = 0$

$- 1 + - 3 = - 4$
$- 1 \cdot - 3 = 3$

${r}^{2} - 4 r + 3 = \left(r - 1\right) \left(r - 3\right)$

$\left(r - 1\right) \left(r - 3\right) = 0$

$r - 1 = 0 \mathmr{and} r - 3 = 0$

$r = 1 \mathmr{and} r = 3$

Mar 13, 2018

Square both sides, then use the quadratic formula to find $r = 1$ and $r = 3$

#### Explanation:

First, square both sides to remove the radical:

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

Next, move all terms to the left hand side to get a quadratic equation:

${r}^{2} - 4 r + 3 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Where $a = 1$, $b = - 4$, and $c = 3$
$x = \frac{4 \pm \sqrt{16 - 12}}{2} \Rightarrow x = \frac{4 \pm 2}{2}$
This give you two solutions, $\textcolor{red}{x = 3}$ and $\textcolor{b l u e}{x = 1}$