find the least a, for which two equations have a common root Could you help me out please. I have two equations: $2x^2-3x+1=0 $  and  $ 2x^2-(a+3)x+3a=0$

I need to find the least $a$ for which these two equations have a common root.
At a first glance I thought it'd be easy, just creating an equation with these two, then creating a function for $a$ and then just a small derivative knowledge. But, unfortunately it seems that it's not as simple as I think, because I've been getting very strange answers.

It'd be wonderful if you could help me out here, just can't concentrate enough maybe.
No derivative knowledge is needed. A common zero of your two polynomials is a root of their difference. So it must be a root of the equation $ax=3a-1$. It is easy to see that $a\ne 0$. So any common root must be equal to $3-1/a$. Substitute in the first equation and solve for $a$.

Because this is homework, we omit the rest of the calculation. But after a while you should get a quadratic in $a$.

Comment: For various reasons, it is nice to put off dividing as long as possible. Since $a\ne 0$, we can rewrite the first equation as $a^2x^2-3a^2x+a^2=0$. Then we can substitute $3a-1$ for $ax$. This yields 
$$2(3a-1)^2 -3a(3a-1)+a^2=0,$$
and then simplification is pleasant and quick.
The resultant of the two polynomials is
$\left| \begin{array} {cccc}
2 &amp; -3 &amp; 1 &amp; 0 \\
0 &amp; 2 &amp; -3 &amp; 1 \\
2 &amp; -a-3 &amp; 3a &amp; 0 \\
0 &amp; 2 &amp; -a-3 &amp; 3a
\end{array} \right| .$

Equating this determinant to $0$ we get $2 (2a-1) (5a-2)=0$, giving $a = \frac 1 2$ and $a = \frac 2 5$.