How do you solve x=sqrt(12x-32)?

Apr 9, 2015

Square both sides of the given equation and then convert into a standard quadratic form and factor.

$x = \sqrt{12 x - 32}$

${x}^{2} = 12 x - 32$

${x}^{2} - 12 x + 32 = 0$

$\left(x - 4\right) \left(x - 8\right) = 0$

$x = 4$
or
$x = 8$

Aug 5, 2018

$x = 4$ and $x = 8$

Explanation:

To get rid of the radical on the right, we can square both sides to get

${x}^{2} = 12 x - 32$

This is beginning to look like a quadratic, but to make it more obvious, we can subtract $12 x$ and add $32$ to both sides to get

${x}^{2} - 12 x + 32 = 0$

To factor this, let's do a little thought experiment:

What two numbers sum up to $- 12$ (middle term) and have a product of the last term ($32$)?

After some trial and error, we arrive at $- 8$ and $- 4$. This allows us to factor this as

$\left(x - 8\right) \left(x - 4\right) = 0$

Setting both factors equal to zero, we get

$x = 4$ and $x = 8$

Hope this helps!