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

2 Answers
Sep 12, 2016

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 6x^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(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!