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.
 A: 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.
A: Note that these equation cannot have two common roots. Because if they did, then one equation is a scalar multiple of the other, in which case, $$\dfrac{2}{2}=\dfrac{-3}{-(a+3)}=\dfrac{1}{3a}$$ The first equality would mean that, $3=a+3 \implies a=0$ which makes the second equality absurd.
So, let $\beta$ be the common root of these two equations. We have, 
$$2\beta^2-3\beta+1=0$$ $$2\beta^2-(a+3)\beta+3a=0$$ Now we use the Cramer's rule to observe that, 
We have that,  $$ \dfrac{\beta^2}{\left| \begin{array}{rr} -3 & 1  \\ -(a+3) & 3a  \end{array} \right| }=\dfrac{\beta}{\left| \begin{array}{rr} 1 & 3a  \\ 2 & 2  \end{array} \right|}=\dfrac{1}{\left| \begin{array}{rr} 2 & 2  \\ -3 & -(a+3)  \end{array} \right|}$$
This gives you, 
$$\dfrac{\beta^2}{-8a+3}=\dfrac{\beta}{2-6a}=\dfrac{1}{-2a}$$
This yields, on eliminating $\beta$, $$\left(\dfrac{2-6a}{-2a}\right)^2=\dfrac{-8a+3}{-2a}$$ This simplifies to the following, $$10a^2-9a+2=0$$ whose roots are $\dfrac{2}{5}$ and $\dfrac{1}{2}$ which implies, the least $a$ is $\dfrac{2}{5}$. So, this completes your answer.
A: The general theory is this: two polynomials have a common root when their resultant is $0$.  There are various ways to calculate that.  The resultant of 
$2 x^2 - 3 x + 1$ and $2 x^2-(a+3) x+3 a$ is $20 a^2-18 a+4 = 2 (5 a - 2) (2 a - 1)$, so there is a common root for $a=2/5$, and $a=1/2$.
A: I think you can just do it in a concrete way. The roots of $2x^2-3x+1=0$ are 
$\frac{1}{2}\mbox{ and } 1.$
By quadratic formula, the roots of $2x^2-(a+3)x+3a=0$ are 
$$\frac{a+3-\sqrt{a^2-18a+9}}{8}\mbox{ and }\frac{a+3+\sqrt{a^2-18a+9}}{8}.$$
If they have common roots, we have the following possibilities:
$$\frac{a+3-\sqrt{a^2-18a+9}}{8}=\frac{1}{2}\mbox{ or }\frac{a+3-\sqrt{a^2-18a+9}}{8}=1,$$
or
$$\frac{a+3+\sqrt{a^2-18a+9}}{8}=\frac{1}{2}\mbox{ or }\frac{a+3+\sqrt{a^2-18a+9}}{8}=1.$$
Each one of them is a quadratic equation in $a$ which can be solved. 
A: The resultant of the two polynomials is
$\left| \begin{array} {cccc}
2 & -3 & 1 & 0 \\
0 & 2 & -3 & 1 \\
2 & -a-3 & 3a & 0 \\
0 & 2 & -a-3 & 3a
\end{array} \right| .$
Equating this determinant to $0$ we get $2 (2a-1) (5a-2)=0$, giving $a = \frac 1 2$ and $a = \frac 2 5$.
