I am working on a question which begins with
The number $\alpha$ is a common root of the equations $x^2+ax+b=0$ and $x^2+cx+d=0$. Given that $a\neq c$, show that $$\alpha=-\frac{b-d}{a-c}$$
This was easy enough to show algebraically by assuming that alpha does satisfy the above two equations. Am I correct in saying that, because I made this assumption, it is sufficient that the root alpha takes on the above form, but it is not necessary? This doesn't seem to fit with the question though, because that would imply that if alpha takes on that value, then the two equations automatically have a common root, however it is not necessary for alpha to take on this value for them to have a common root. In that case, I have not shown that alpha being a common root of the equations implies that alpha must take on this value, I have shown that if alpha does take. On this value, then the two equations have a common root... I'm sorry if this is very confusing! I am trying to explain myself as best as I can...
The second part of the question is
Hence, or otherwise, show that the equations have at least one common root if and only if $$(b-d)^2-a(b-d)(a-c)+b(a-c)^2=0$$
Since I am not sure whether the first statement for the value of alpha is necessary or sufficient, I am having trouble deciding how far I can deal with the second part of the question using the result from the first part...
Thank you in advance for any guidance. If anyone has a link to any particularly good resource that will help with using necessary/sufficient and if/iff I would be very grateful.
Assume for now that $a\ne c$. Then you’ve shown that if if the two equations have a common root $\alpha$, then that common root must be
$$-\frac{b-d}{a-c}\;.\tag{1}$$
You’ve not shown, however, that the expression in $(1)$ always is a common root of the two equations. That is, you’ve shown that having $\alpha$ equal to $(1)$ is necessary, but not that it’s sufficient. And in fact it’s not sufficient. Consider the equations $x^2=0$ and $x^2+x+1=0$, with $a=b=0$ and $c=d=1$: then
$$-\frac{b-d}{a-c}=-\frac{0-1}{0-1}=-1\;,$$
but $-1$ isn’t a root of either equation.
To find a sufficient condition, substitute the expression $(1)$ for $x$ in each of the equations; in the first you get
$$\frac{(b-d)^2}{(a-c)^2}-\frac{a(b-d)}{a-c}+b$$
on the left-hand side, so in order for $(1)$ to be a root of $x^2+ax+b=0$, we must have
$$\frac{(b-d)^2}{(a-c)^2}-\frac{a(b-d)}{a-c}+b=0\;.\tag{2}$$
Similarly, from the second equation we find that we must have
$$\frac{(b-d)^2}{(a-c)^2}-\frac{c(b-d)}{a-c}+d=0\;.\tag{3}$$
Multiplying $(2)$ through by $(a-c)^2$ yields
$$(b-d)^2-a(b-d)(a-c)+b(a-c)^2=0\;,\tag{4}$$
which should look familiar. Doing the same thing to $(3)$ yields
$$(b-d)^2-c(b-d)(a-c)+d(a-c)^2=0\;;\tag{5}$$
with a bit of algebra you can show that $(4)$ and $(5)$ are equivalent, in the sense that each is true if and only if the other is true. (HINT: Solve them for $\frac{(b-d)^2}{a-c}$.) Thus, we need only consider $(4)$.
If $(4)$ holds, and $a\ne c$, then $(2)$ and $(3)$ hold, and $(1)$ is a root of the first equation. What does $(4)$ imply when $a=c$? Does it still ensure that the two equations have a common root?