Question

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

Answer 1

x = 1 only

Explanation:

rewrite as `sqrt(3x^2 - 12x + 9)` = 3x - 3
square both sides `3x^2` - 12x + 9 = `9x^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.

Answer 2

Just to clarify what the previous contributor said.

Explanation:

Isolate the `sqrt` on one side of the equation:

`sqrt(3x^2 - 12x + 9) = 3x - 3`

`(sqrt(3x^2 - 12x + 9))^2 = (3x -3)^2`

`3x^2 - 12x + 9 = 9x^2 - 18x + 9`

`-6x^2 + 6x = 0`

`-6x(x - 1) = 0`

`x = 0 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!