Step by step please
Solve
$\displaystyle \frac{3}{(y-1)^{2}}+\frac{3}{2(y-1)^{2}}=2$
I got to
$\displaystyle 4=(y-1)^{2}$ or $\displaystyle 4=y^{2}-2y+1$
Multiply throughout by $\displaystyle 2(y-1)^2$ to clear the denominator
$\displaystyle 2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2$
Simplify
$\displaystyle 6 + 3 = 4(y-1)^2$
You could expand and use the quadratic formula but completing the square is more appropriate here IMO
Divide by 4
$\displaystyle (y-1)^2 = \frac{9}{4} = \left(\frac{3}{2}\right)^2$
Take the square root of both sides and don't forget the $\displaystyle \pm$ sign
another way
x every term by 2 you have
$\displaystyle
\frac{9}{(y-1)^2} = 4
\Rightarrow
4(y-1)^2 = 9
\Rightarrow
(y-1)^2 = \frac{9}{4}
$
take the square of both sides
$\displaystyle
y-1 = \frac{3}{2}\Rightarrow
y = \frac{3}{2} + \frac{2}{2} = \frac{5}{2}
$
Do you see the error you made here, Mukilab?
You could have cancelled terms on the left, you did cancel terms
in the middle but you dropped a term on the right.
Did you understand what $\displaystyle e^{i*pi}$ meant when he mentioned + and - in relation to the square roots?
There's no problem,
no-one is judging,
but you learn from seeing your errors and correcting them.
Now, there are 2 answers,
since (3)(3)=9 and (-3)(-3)=9 for example.
Of course, they are the same thing really.
3(-3) means subtract 3 three times,
while -3(-3) means the opposite of that which is add 3 three times.
So, square roots have 2 answers.