$ax^2 +bx +cm =0$ , $bx^2 + cx +am =0$ and $cx^2 + ax +bm=0$ are three quadratic equations in $x$ , $a,b ,c$ are real numbers and $m$ is a positive real , find the possible numerical values of $m$ so that atleast one of these equations has a real root.

How do I attempt such a question? What is the intuition behind this?

I don't get where to start. Can someone help me out?

I got $b^2 \ge 4acm$, $c^2 \ge 4abm$, $a^2 \ge 4bcm$ but what do I do with these? Atleast one of them has to be true? Is there something else I should try?

Thanks in advance.

  • $\begingroup$ List the three inequalities about the discriminant and see what you can deduce. $\endgroup$
    – justadzr
    Aug 23 at 3:35
  • $\begingroup$ I have tried that , but it did not lead me to anything. $\endgroup$ Aug 23 at 3:37
  • $\begingroup$ Then you could type out what you did and then others will give you some further hints. And try to use Latex, though the expressions in your question are readable. $\endgroup$
    – justadzr
    Aug 23 at 3:37

Let's look at the case where all of them have no real roots. This means $b^2 - 4acm < 0, c^2 - 4abm < 0, a^2 - 4bcm < 0\implies a^2+b^2+c^2 < m(4ab+4bc+4ca)\implies m > \dfrac{a^2+b^2+c^2}{4ab+4bc+4ca}$. Thus if $0 < m \le \dfrac{a^2+b^2+c^2}{4ab+4bc+4ca}$, then there is at least one of the inequalities above which is non-negative, and in such case the corresponding equation would have a real root.

Note: You can show first that $ab > 0$. This comes from $4abm > c^2 \ge 0 \implies ab > 0$ since $m > 0$. Hence $ab > 0$, and similarly $bc > 0, ca > 0$. So $ab+bc+ca > 0$.

  • $\begingroup$ That's assuming $\,ab+bc+ca \gt 0\,$ which is not given in the problem. $\endgroup$
    – dxiv
    Aug 23 at 5:43
  • $\begingroup$ How can we say that if $0<m\leq ...$ atleast one of the inequalities above which is non-negative? $\endgroup$ Aug 23 at 6:39
  • $\begingroup$ @LalitTolani: it comes from a fairly straightforward contrapositive argument. If $A \implies B$ is true, then it is also true that $ \bar{B} \implies \bar{A}$. Here the $\bar{X}$ means not $X$, and the $A$ is all three inequalities are true, and the $B$ is the $m >...$ $\endgroup$
    – Wang YeFei
    Aug 23 at 6:46
  • 3
    $\begingroup$ But , if discriminants are 2 , -3 , -5 , their sum is negative but atleast one of them has a solution. $\endgroup$ Aug 23 at 6:48
  • $\begingroup$ @WangYeFei "You can show first that $ab \gt 0$" No, you cannot show that. You don't get to choose $\,a,b,c\,$, those are given and you must find the values $\,m \gt 0\,$ which satisfy the problem (if any such exist at all). $\endgroup$
    – dxiv
    Aug 23 at 17:52

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