# How do you solve the equation: x=sqrt(3x+40)?

May 28, 2018

$x = 8$

#### Explanation:

$x = \sqrt{3 x + 40}$

Over reals the radical sign refers to the principal square root, so squaring both sides may introduce extraneous solutions.

${x}^{2} = 3 x + 40$

${x}^{2} - 3 x - 40 = 0$

$\left(x - 8\right) \left(x + 5\right) = 0$

$x = 8 \mathmr{and} x = - 5$

The principal square root won't ever be $- 5$ so we reject that as extraneous.

$x = 8$

Check:

 sqrt{3(8)+40}=sqrt{64}=8 quad sqrt