How do you solve $x = \sqrt{12 x - 32}$ $x=sqrt(12x-32)$ ?

Question

2646 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-09T17:17:32

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

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

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

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

$(x - 4) (x - 8) = 0$ $(x-4)(x-8) = 0$

$x = 4$ $x=4$
or
$x = 8$ $x=8$

Answer 2 · 2018-08-05T01:46:47

$x = 4$ $x=4$ and $x = 8$ $x=8$

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

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

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

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

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

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

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

$(x - 8) (x - 4) = 0$ $(x-8)(x-4)=0$

Setting both factors equal to zero, we get

$x = 4$ $x=4$ and $x = 8$ $x=8$

Hope this helps!

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