Condition for exactly one root being common b/w two quadratic equations 
Question Statement:-
If the equations $ax^2+2bx+c=0$ and $a_1x^2+2b_1x+c_1=0$ have one and only one root common, then prove that $b^2-ac$ and $b_1^2-a_1c_1$ are perfect squares.


Attempt at a solution:-
Let the common root be $\alpha$. Then $x=\alpha$ must satisfy both the quadratic equations. Hence, we have
$a\alpha^2+2b\alpha+c=0\tag{1}$
$a_1\alpha^2+2b_1\alpha+c_1=0\tag{2}$
On solving $(1)$ and $(2)$, we get
$$\dfrac{\alpha^2}{2bc_1-2b_1c}=\dfrac{-\alpha}{ac_1-a_1c}=\dfrac{1}{2ab_1-2ba_1}$$
From this, we get
$$(a_1c-ac_1)^2=(2ab_1-2a_1b)(2bc_1-2b_1c)$$
On expanding and manipulating this I couldn't get anywhere close to what the question needs me to prove.
If you can think of an intuitive solution using graphs of the quadratic equations then it would be helpful too.
 A: This answer assumes $a,b,c$ are integers (and of course, if they are rational, one can multiply both sides of the equation so that this holds). Write the equations as
$$a(x-r)(x-s)=ax^2-a(r+s)x+ars=0\\a_1(x-r)(x-t)=a_1x^2-a_1(r+t)x+a_1rt=0$$
Of course, $r \neq s,t$. These imply:
\begin{align}
&-a(r+s)=2b&&ars=c\\
&-a_1(r+t)=2b_1&&a_1rt=c_1
\end{align}
Then:
\begin{align}
b^2-ac&=\frac{a^2(r+s)^2}{4}-a^2rs\\
&=\frac{a^2}{4}(r^2+2rs+s^2-4rs)\\
&=\frac{a^2}{4}(r^2-2rs+s^2)\\
&={\left(\frac{a(r-s)}{2}\right)}^2
\end{align}
Notice that because $a(r+s)$ is even ($-2b)$, $a$ is even or $r,s$ are the same parity. Whatever the case, $a(r-s)$ will be even, so the fraction above is an integer and we are done.
The case ${b_1}^2-a_1c_1$ is treated similarly.
A: $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b_1\pm\sqrt{b_1^2-4a_1c_1}}{2a_1}$$
$$\sqrt{b^2-4ac}+-\sqrt{b_1^2-4a_1c_1}\in Q$$
When you square it you will get
$$\sqrt{(b^2-4ac)(b_1^2-4a_1c_1)}\in Q$$
So
$$b^2-4ac=p^2m,\qquad b_1^2-4a_1c_1=q^2m$$
$$(p+-q)\sqrt m \in Q$$
$$\sqrt m \in Q$$
