# How do you solve sqrt(26x-39)=x+5?

Apr 16, 2015

We can square both sides to get

${\left(\sqrt{26 x - 39}\right)}^{2} = {\left(x + 5\right)}^{2}$

$26 x - 39 = {x}^{2} + 10 x + 25$
(Used the Identity color(blue)((a + b)^2 = a^2 + 2ab + b^2

Transposing the terms from the left hand side to the right, we get:

$0 = {x}^{2} + 10 x - 26 x + 25 + 39$

$0 = {x}^{2} - 16 x + 64$

${x}^{2} - 16 x + 64 = 0$

${\left(x - 8\right)}^{2} = 0$

 color(green)(x = 8

Left Hand Side = $\sqrt{26 \cdot 8 - 39} = \sqrt{169} = 13$
Right Hand Side = $8 + 5 = 13$