Find the range of $m$ for which atleast one of the following equations $ax^2+bx+cm=0,bx^2+cx+am=0,cx^2+ax+bm=0$ have real roots.

Question

Let $a,b,c$ and $m\in R^+$.Find the range of $m$ for which atleast one of the following equations $ax^2+bx+cm=0,bx^2+cx+am=0,cx^2+ax+bm=0$ have real roots.

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Either one or two or all of the three equations $ax^2+bx+cm=0,bx^2+cx+am=0,cx^2+ax+bm=0$ have the real roots.

But i do not know how to find the range of $m.$Any help will be appreciated.Thanks.

Answer 1

Claim: $$m \leq \dfrac{1}{4}$$

For if all of them do not have real roots then : $a^2-4bcm < 0, b^2-4acm < 0, c^2-4abm < 0 \Rightarrow a^2+b^2+c^2 < 4m(ab+bc+ca) \Rightarrow m > \dfrac{a^2+b^2+c^2}{4ab+4bc+4ca} \geq \dfrac{ab+bc+ca}{4ab+4bc+4ca} = \dfrac{1}{4}$. Thus if $m \leq \dfrac{1}{4}$, then at least one equation has real roots.

Answer 2

At least one of the three discriminants has to be positive:

$b^2-4acm\geq0$

$c^2-4bam\geq0$

$a^2-4cbm\geq0$

Equivalently:

$\frac{b^2}{4ac}\geq m$

$\frac{a^2}{4ba}\geq m$

$\frac{c^2}{4bc}\geq m$.

One of these is true as long as $m$ is less or equal to the maximum of the three left hand sides.

So let $M=\max\{\frac{b^2}{4ac},\frac{a^2}{4ba},\frac{c^2}{4bc}\}$.

Since $m\geq0$ we must have $0\leq m\leq M$.