# How do you solve sqrt{3x^{2} - 12x + 9} + 3= 3x?

Sep 12, 2016

x = 1 only

#### Explanation:

rewrite as $\sqrt{3 {x}^{2} - 12 x + 9}$ = 3x - 3
square both sides $3 {x}^{2}$ - 12x + 9 = $9 {x}^{2}$ - 18x + 9
collect like terms and simplify 6${x}^{2}$ - 6x = 0
divide by 6 ......... ${x}^{2}$ - x = 0
factorise gives x(x - 1) = 0
so possibly x = 0 or x = 1
check in given equation if valid, if x = 0 then 3 = 0 not possible
so x = 0 is not a valid solution.
check in given equation if valid, if x = 1 then 3 = 3 is possible
so x = 1 is a valid solution.

Sep 12, 2016

Just to clarify what the previous contributor said.

#### Explanation:

Isolate the sqrt on one side of the equation:

$\sqrt{3 {x}^{2} - 12 x + 9} = 3 x - 3$

${\left(\sqrt{3 {x}^{2} - 12 x + 9}\right)}^{2} = {\left(3 x - 3\right)}^{2}$

$3 {x}^{2} - 12 x + 9 = 9 {x}^{2} - 18 x + 9$

$- 6 {x}^{2} + 6 x = 0$

$- 6 x \left(x - 1\right) = 0$

$x = 0 \mathmr{and} 1$

However, checking in the original equation, $x = 0$ is extraneous since it doesn't satisfy the original equation. The only actual solution is $x = 1$.

Hopefully this helps!