Quadratic Equations - Condition for real roots

[Saitama]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I don't really know where to start. If the quadratic has real roots, the discriminant is greater than or equal to zero. This would give me three inequations for each quadratic but how should I deal with "at least" one of the equation having real roots?

 

[Mandelbroth]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I don't really know where to start. If the quadratic has real roots, the discriminant is greater than or equal to zero. This would give me three inequations for each quadratic but how should I deal with "at least" one of the equation having real roots?

$$b^2-4acm \geq 0 \\ c^2-4bam \geq 0 \\ a^2-4cbm \geq 0$$
I'm confused as to your meaning in the first sentence. Was what you meant equivalent to $\{a,b,c,m\}\subset\mathbb{R}_{+}$?

 

[Mark44]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I don't really know where to start. If the quadratic has real roots, the discriminant is greater than or equal to zero. This would give me three inequations for each quadratic but how should I deal with "at least" one of the equation having real roots?

Break the problem into three cases.
1. Assume that the first equation has real roots.
2. Assume that the second equation has real roots.
3. Assume that the third equation has real roots.

For each case, what can you say about the discriminant, which will have a different form for each case? I haven't worked the problem, but this is how I would approach it.

 

[HallsofIvy]

Adding to what Mandelbroth said, $b^2- 4acm\ge 0$ is equivalent to $m\le b^2/4ac$, $c^2- 4bam\ge 0$ is equivalent to $m\le c^2/4ab$, and $a^2- 4cbm\ge 0$ is equivalent to $m\le a^2/4bc$. If you do not know any relations among a, b, and, c, the most you can say is that at least one of those must be true.

 

[I like Serena]

One of the 3 inequalities with $m$ on the left side must have the greatest value on the right hand side.
That would be the one with the greatest value in the numerator.

Suppose $a$ has the greatest value.
So $a \ge b$ and $a \ge c$.
Then what can you say about this greatest right hand value for sure?

 

[haruspex]

Break the problem into three cases.
1. Assume that the first equation has real roots.
2. Assume that the second equation has real roots.
3. Assume that the third equation has real roots.

For each case, what can you say about the discriminant, which will have a different form for each case? I haven't worked the problem, but this is how I would approach it.

I suggest it would be easier to invert the logic: suppose none of them has real roots. Then you can look for a way of combining the inequalities that eliminates a, b and c. For an m that satisfies the condition that emerges, there likely exists a triple (a, b, c) s.t. none of the equations has real roots (but you'll need to show the triple does exist). Finally, you need to show that for all other m no such triple exists.

 

[Saitama]

One of the 3 inequalities with $m$ on the left side must have the greatest value on the right hand side.
That would be the one with the greatest value in the numerator.

Suppose $a$ has the greatest value.
So $a \ge b$ and $a \ge c$.
Then what can you say about this greatest right hand value for sure?

I have three inequalities,
$m\le b^2/4ac, m\le c^2/4ab$ and $m\le a^2/4bc$.

For $a\geq b$ and $a \geq c$, $a^2/4bc$ has the greatest value. What next?

I suggest it would be easier to invert the logic: suppose none of them has real roots. Then you can look for a way of combining the inequalities that eliminates a, b and c. For an m that satisfies the condition that emerges, there likely exists a triple (a, b, c) s.t. none of the equations has real roots (but you'll need to show the triple does exist). Finally, you need to show that for all other m no such triple exists.

When none of them has real roots, the inequalities are
$m> b^2/4ac, m > c^2/4ab$ and $m > a^2/4bc$
Adding the inequalities,
$$12m>\frac{a^3+b^3+c^3}{abc}$$
What should I do next? I can find the minimum value of RHS of inequality but I don't think that would help.

 

[I like Serena]

I have three inequalities,
$m\le b^2/4ac, m\le c^2/4ab$ and $m\le a^2/4bc$.

For $a\geq b$ and $a \geq c$, $a^2/4bc$ has the greatest value. What next?

With $a\geq b$ and $a \geq c$, you get that $a^2 \ge bc$, ...

 

[Saitama]

With $a\geq b$ and $a \geq c$, you get that $a^2 \ge bc$, ...

The only thing I understand is that the minimum value of $a^2/(4bc)$ is 1/4.

 

[verty]

The point is, which m satisfies all inequalities? For m to be greater than all 3, it must be greater than the greatest.

 

[haruspex]

When none of them has real roots, the inequalities are
$m> b^2/4ac, m > c^2/4ab$ and $m > a^2/4bc$
Adding the inequalities,
$$12m>\frac{a^3+b^3+c^3}{abc}$$
What should I do next? I can find the minimum value of RHS of inequality but I don't think that would help.

Adding them didn't get rid of a, b and c. Try something else... something, shall we say, more productive

 

[verty]

Oh, I didn't understand what was meant by "independent of a,b,c", I see now that it means "find a range of m that will guarantee one of these has real solutions given a,b,c > 0".

 

[I like Serena]

The only thing I understand is that the minimum value of $a^2/(4bc)$ is 1/4.

Yes. So without any knowledge of a, b, and c, you can still guarantee that there will be a solution as long as $0<m \le \,^1\!\!/\!_4$.

 

[Saitama]

Adding them didn't get rid of a, b and c. Try something else... something, shall we say, more productive

Multiply the inequalities?
That gives $$m>\frac{1}{4}$$
I think this means that m should be less than or equal to 1/4 so that at least one of them have real roots. So I guess this is enough for the question?

Yes. So without any knowledge of a, b, and c, you can still guarantee that there will be a solution as long as $0<m \le \,^1\!\!/\!_4$.

Yep!

Both the solutions are interesting, thank you both ILS and haruspex!

 

[haruspex]

m should be less than or equal to 1/4 so that at least one of them have real roots. So I guess this is enough for the question?

You've shown that no matter what a, b and c are, if there are no roots then m > 1/4.
$\forall a \in\Re^+ \forall b \in\Re^+ \forall c \in\Re^+, m \in [0\frac 14] \Rightarrow \exists (a root)$
Equivalently
$m \in [0\frac 14] \Rightarrow (\forall a \in\Re^+ \forall b \in\Re^+ \forall c \in\Re^+, \exists (a root))$
The question (I think) asks for the complete range R of m for which, no matter what a, b and c are, there are guaranteed some roots.
$m \in R \Leftrightarrow (\forall a \in\Re^+ \forall b \in\Re^+ \forall c \in\Re^+, \exists (a root))$