# How do you solve the equation sqrt(5r-1)=r-5?

Oct 26, 2017

$r = 13$

#### Explanation:

As we have $\sqrt{5 r - 1} = r - 5$, domain of $r$ is $\left[\frac{1}{5} , \infty\right)$

Squaring both sides in $\sqrt{5 r - 1} = r - 5$ we get

$5 r - 1 = {\left(r - 5\right)}^{2}$

or $5 r - 1 = {r}^{2} - 10 r + 25$

or ${r}^{2} - 15 r + 26 = 0$

or ${r}^{2} - 13 r - 2 r + 26 = 0$

or $r \left(r - 13\right) - 2 \left(r - 13\right) = 0$

or $\left(r - 2\right) \left(r - 13\right) = 0$

i.e. $r = 2$ or $13$

Observe that both values are within domain but $r = 2$ results in $\sqrt{5 r - 1} = \sqrt{9} = 3 \ne 2 - 5$, as it takes negative square root of $\sqrt{5 r - 1}$ for equality to hold.

Hence, answer is $r = 13$.