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 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(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!