# How do you solve sqrt(3x)+8=x+2?

Mar 11, 2016

$x = \left\{3 , 12\right\}$

#### Explanation:

$\sqrt{3 x} + 8 = x + 2$
$\sqrt{3 x} = x + 2 - 8$
$\sqrt{3 x} = x - 6$
${\left(\sqrt{3 x}\right)}^{2} = {\left(x - 6\right)}^{2}$
$3 x = {x}^{2} - 12 x + 36$
${x}^{2} - 12 x - 3 x + 36 = 0$
${x}^{2} - 15 x + 36 = 0$
$\left(x - 12\right) \left(x - 3\right) = 0$
$\text{if (x-12)=0 then x=12}$
$\text{if (x-3)=0 then x=3}$
$x = \left\{3 , 12\right\}$

Mar 11, 2016

3 and 12

#### Explanation:

$\sqrt{3 x} + 8 = x + 2$
Isolate the radical term.
$\sqrt{3 x} = x - 6$
Square both sides:
$3 x = {\left(x - 6\right)}^{2} = {x}^{2} - 12 x + 36$
${x}^{2} - 15 x + 36 = 0$
Find 2 numbers (real roots) knowing sum (15 = -b) and product (c = 36). They are: 3 and 12.