# Condition for system of quadratic equations to have atleast one solution

$$ax^2 +bx +cm =0$$ , $$bx^2 + cx +am =0$$ 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?

• List the three inequalities about the discriminant and see what you can deduce. Aug 23 at 3:35
• I have tried that , but it did not lead me to anything. Aug 23 at 3:37
• 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. 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 $$ab+bc+ca > 0$$.

• That's assuming $\,ab+bc+ca \gt 0\,$ which is not given in the problem.
– dxiv
Aug 23 at 5:43
• How can we say that if $0<m\leq ...$ atleast one of the inequalities above which is non-negative? Aug 23 at 6:39
• @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 $m >...$ Aug 23 at 6:46
• But , if discriminants are 2 , -3 , -5 , their sum is negative but atleast one of them has a solution. Aug 23 at 6:48
• @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).
– dxiv
Aug 23 at 17:52