Question 30038
3x^2 - 2x - 3 = 0   ----(1)
Multiplying by 3
9x^2-6x-9 =0 ----(2)  (zero mulitplied by anything is zero)
(3x)^2- 2X(3x)-9= 0
That is (3x)^2- 2X(3x)X(1)= 9
[The left hand side is like (a)^2 -2ab with a =(3x) and b = 1. we require b^2 to make it a perfect square: (a)^2 -2ab +b^2 = (a-b)^2 ]
Adding b^2 = 1^2 = 1 on both the sides
(3x)^2- 2X(3x)X (1) +1 = 9 +1
(3x-1)^2 = 10
Taking the sqrt on both the sides
(3x-1) = +or- sqrt(10)
3x = 1+or- sqrt(10)
x = (1/3)times[1+or- sqrt(10)]
Answer: x = (1/3)(1+sqrt10) and x= (1/3)(1-sqrt10)
Verification: x = (1/3)(1+sqrt10) in (1) implies
LHS = 3x^2 - 2x - 3 
= 3X(1/9)(1+10+2sqrt10)-2X(1/3)X(1+sqrt10)-3
=(1/3)[11+2sqrt10-2(1+sqrt10)]-3
=(1/3)[11+2sqrt10-2-2sqrt10)]- 3
=(1/3)(11-2) - 3
=(1/3)X(9) - 3
=3-3 = 0 =RHS
Therefore x = (1/3)(1+sqrt10)  is correct
Since surd roots occur in conjuate pairs in equations 
there is no need for us to test the validity of the other root.
Note: Why did we multiply through out by 3 in the beginning?
Abs: To make the first term a perfect square 
to play the role of a in the perfect square (a-b)^2