# How do you solve 2x=sqrt[12x-5] and find any extraneous solutions?

Nov 9, 2017

Start by squaring both sides...

#### Explanation:

...giving $4 {x}^{2} = 12 x - 5$

...now you can subtract 12x and add 5 to both sides, giving:

$4 {x}^{2} - 12 x + 5 = 0$

...this is a quadratic equation with coefficients 4, -12, and 5.

Remember your forumula, if $a {x}^{2} + b x + c = 0$, then your roots are:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

...and for coefficients a = 4, b = -12, c = 5, your roots are:

2.5 and 0.5

GOOD LUCK