# How do you solve for k in d + 3 = sqrt(2k – 5)?

##### 1 Answer
Feb 29, 2016

$k = \frac{{d}^{2} + 6 d + 14}{2}$

#### Explanation:

$d + 3 = \sqrt{2 k - 5}$

we can exchange the order off the factors:

$\sqrt{2 k - 5} = d + 3$

Now square both sides of the equation. sqrt(2k-5) is a positive number so we must add the constriction d+3>=0. Otherwise we would add a false solution:

${\left(\sqrt{2 k - 5}\right)}^{2} = {\left(d + 3\right)}^{2} \mathmr{and} d + 3 \ge 0$

$2 k - 5 = {d}^{2} + 6 d + 9 \mathmr{and} d \ge - 3$

$2 k = {d}^{2} + 6 d + 14 \mathmr{and} d \ge - 3$

$k = \frac{{d}^{2} + 6 d + 14}{2} \mathmr{and} d \ge - 3$