# How do you solve for a in sqrta + 9 = b + 3 ?

Mar 15, 2018

While this equation can't actually be solved to give an integer value for $a$, it can be rearranged, though it doesn't provide much use:
$\sqrt{a} = b + 3 - 9$
$\sqrt{a} = b - 6$
$a = \sqrt{a} \left(b - 6\right)$