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


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.

  • $\begingroup$ I suppose $a,b,c$ are integers? $\endgroup$ Nov 8, 2016 at 14:50
  • $\begingroup$ @Fimpellizieri - Why is it needed to be supposed that $a,b,c$ are integers, for your reference the book does not mention so. $\endgroup$
    – user350331
    Nov 8, 2016 at 16:23
  • 1
    $\begingroup$ @user350331 The equations $x^2-6x+5$ and $\sqrt\pi x^2 - 4\sqrt\pi x + 3\sqrt\pi$ both have the root $x=1$ and no other root in common, but $(2\sqrt\pi)^2 - \sqrt\pi \cdot 3\sqrt\pi = \pi$ is not a perfect square. Your book may mention several pages earlier that you will only be using integers in this chapter, or there may be some other indication of what is intended. $\endgroup$
    – David K
    Nov 8, 2016 at 16:41
  • $\begingroup$ @DavidK- Oh yeah it does tell "to consider the coefficients to be integers unless stated otherwise". Thanks for that I had been solving my questions without considering that and now only did I think about it. $\endgroup$
    – user350331
    Nov 8, 2016 at 16:44

2 Answers 2


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


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}


\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.

  • $\begingroup$ Why is it necessary to assume that the coefficients are integer in both the quadratic equations. $\endgroup$
    – user350331
    Nov 8, 2016 at 16:37
  • $\begingroup$ Because otherwise we may have that $b^2-ac$ is not an integer (so it can't be a square). $\endgroup$ Nov 8, 2016 at 17:11

$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b_1\pm\sqrt{b_1^2-4a_1c_1}}{2a_1}$$ $$\sqrt{b^2-4ac}+-\sqrt{b_1^2-4a_1c_1}\in Q$$ When you square it you will get $$\sqrt{(b^2-4ac)(b_1^2-4a_1c_1)}\in Q$$ So $$b^2-4ac=p^2m,\qquad b_1^2-4a_1c_1=q^2m$$ $$(p+-q)\sqrt m \in Q$$ $$\sqrt m \in Q$$

  • $\begingroup$ Doesn't the first line mean that you are equating both the roots of both the equation. $\endgroup$
    – user350331
    Nov 8, 2016 at 16:35
  • $\begingroup$ @user350331 I mean one of the combinations + or -, isn't matter which one it is $\endgroup$ Nov 8, 2016 at 19:18

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