# complex algebra

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• Jun 25th 2010, 10:30 PM
hooke
complex algebra
Given that the quadratic equation $(m^2-3m-4)x^2+(m^2+2)x+12=0$ has real roots $\alpha$ and $\beta$ .Find the set of values of m such that $\alpha<-1<\beta$

Attempt:

$\alpha+\beta=-[\frac{m+2}{m^2-3m-4}]$

$\alpha\beta=\frac{12}{m^3-3m-4}$

Expressing beta in terms of m,

$\beta=\frac{(m+2)\pm \sqrt{(m+2)^2-4(4+3m-m^2)(-12)}}{2(4+3m-m^2)}$

Then since \beta>-1, i have to find the set of values of m such that the long and boring stuff i found above is >-1, but how do i solve that inequality? Am i even on the right track?

• Jun 26th 2010, 02:14 AM
Ackbeet
Hmm. I think your thoughts are moving in the right direction. However, you have a lot of sign errors, as well as exponent errors. The addition of alpha and beta needs to be divided by 2. The use of the quadratic formula should give you both alpha and beta. You should be able, because of the inequality constraint, to assign the + solution to one variable, and the - to the other, where the + and the - correspond to the sign difference in the quadratic formula between the two roots. Which one is which, do you think?
• Jun 26th 2010, 05:54 AM
hooke
Quote:

Originally Posted by Ackbeet
Hmm. I think your thoughts are moving in the right direction. However, you have a lot of sign errors, as well as exponent errors. The addition of alpha and beta needs to be divided by 2. The use of the quadratic formula should give you both alpha and beta. You should be able, because of the inequality constraint, to assign the + solution to one variable, and the - to the other, where the + and the - correspond to the sign difference in the quadratic formula between the two roots. Which one is which, do you think?

thanks, but could u pls indicate next to my working the errors that i made

don really get what u mean sorry..
• Jun 26th 2010, 06:23 AM
running-gag
Hi

Let $f(x) = (m^2-3m-4)x^2+(m^2+2)x+12$

When $m^2-3m-4 > 0$ which means $m < -1$ or $m > 4$ the representative curve of f is a parabola oriented upwards (limit $+\infty$ at both sides)

Consider a condition on f(-1) to get $\alpha < -1 < \beta$

Then do the same when $m^2-3m-4 < 0$
• Jun 26th 2010, 06:47 AM
hooke
Quote:

Originally Posted by running-gag
Hi

Let $f(x) = (m^2-3m-4)x^2+(m^2+2)x+12$

When $m^2-3m-4 > 0$ which means $m < -1$ or $m > 4$ the representative curve of f is a parabola oriented upwards (limit $+\infty$ at both sides)

Consider a condition on f(-1) to get $\alpha < -1 < \beta$

Then do the same when $m^2-3m-4 < 0$

thanks but i don understand what u meant by considering a condition on f(-1), could you explain a little on that?
• Jun 26th 2010, 10:15 AM
running-gag
When the parabola is oriented upwards, if f(-1) < 0 due to the intermediate value theorem there exists $\alpha < -1$ such that $f(\alpha) = 0$ and $\beta > -1$ such that $f(\beta) = 0$
• Jun 26th 2010, 05:46 PM
hooke
thanks Runninggag, but is there a more elementary solution to this problem ,preferably an algebraic solution since this is a high school question?
• Jun 26th 2010, 06:39 PM
Ackbeet
Here are my computations:

Given $(m^{2}-3m-4)x^{2}+(m^{2}+2)x+12=0.$

Assume the given equation has real roots $\alpha$ and $\beta$.

Find all $m$ such that $\alpha<-1<\beta$.

If we write the original equation as

$x^{2}+\frac{m^{2}+2}{m^{2}-3m-4}\,x+\frac{12}{m^{2}-3m-4}=0$, and compare with

$(x-\alpha)(x-\beta)=x^{2}-\beta x-\alpha x+\alpha\beta
=x^{2}-(\alpha+\beta)x+\alpha\beta=0$
, then we see that

$-(\alpha+\beta)=\frac{m^{2}+2}{m^{2}-3m-4}$ and

$\alpha\beta=\frac{12}{m^{2}-3m-4}$.

Check these expressions against yours to see where you might have gone wrong.

I would probably, at this point, solve these two equations for $\alpha$ and $\beta$ in terms of $m$. What do you get for that?
• Jun 26th 2010, 06:53 PM
hooke
Quote:

Originally Posted by Ackbeet
Here are my computations:

Given $(m^{2}-3m-4)x^{2}+(m^{2}+2)x+12=0.$

Assume the given equation has real roots $\alpha$ and $\beta$.

Find all $m$ such that $\alpha<-1<\beta$.

If we write the original equation as

$x^{2}+\frac{m^{2}+2}{m^{2}-3m-4}\,x+\frac{12}{m^{2}-3m-4}=0$, and compare with

$(x-\alpha)(x-\beta)=x^{2}-\beta x-\alpha x+\alpha\beta
=x^{2}-(\alpha+\beta)x+\alpha\beta=0$
, then we see that

$-(\alpha+\beta)=\frac{m^{2}+2}{m^{2}-3m-4}$ and

$\alpha\beta=\frac{12}{m^{2}-3m-4}$.

Check these expressions against yours to see where you might have gone wrong.

I would probably, at this point, solve these two equations for $\alpha$ and $\beta$ in terms of $m$. What do you get for that?

thanks, yeah there are a couple of typos in my OP so i got

$\alpha^2(m^2-3m-4)+(m^2+2)\alpha+12=0$

Solve for alpha using the quadratic formula, as usual i got a pretty long stuff.

What's the next step?
• Jun 27th 2010, 05:23 AM
Dinkydoe
Quote:

thanks, yeah there are a couple of typos in my OP so i got

Solve for alpha using the quadratic formula, as usual i got a pretty long stuff.

What's the next step?
We have $\alpha,\beta = \frac{-(m^2+2)\pm \sqrt{m^4-44m^2+144m+196}}{2(m^2-3m-4)}$

Thus if we want to have $\alpha < -1 < \beta$ then we can impose 2 conditions:

1. $\Delta(f) = m^4-44m^2+144m+196> 0$

2. $\alpha =\frac{-(m^2+2) - \sqrt{\Delta(f)})}{2(m^2-3m-4)}< -1 <\frac{-(m^2+2) + \sqrt{\Delta(f)})}{2(m^2-3m-4)}=\beta$

Or, $-(m^2+2)-\sqrt{\Delta(f)}<- 2(m^2-3m-4) < -(m^2+2)+\sqrt{\Delta(f)}$

or, $-\sqrt{\Delta(f)} < -m^2+6m+10 < \sqrt{\Delta(f)}$

And this inequality can be rewritten to:

$-\Delta(f) < (-m^2+6m+10)^2 < \Delta(f)$

Now we're left with finding m such that $\Delta(f) > (-m^2+6m+10)^2$
• Jun 27th 2010, 06:13 AM
hooke
Quote:

Originally Posted by Dinkydoe
We have $\alpha,\beta = \frac{-(m^2+2)\pm \sqrt{m^4-44m^2+144m+196}}{2(m^2-3m-4)}$

Thus if we want to have $\alpha < -1 < \beta$ then we can impose 2 conditions:

1. $\Delta(f) = m^4-44m^2+144m+196> 0$

2. $\alpha =\frac{-(m^2+2) - \sqrt{\Delta(f)})}{2(m^2-3m-4)}< -1 <\frac{-(m^2+2) + \sqrt{\Delta(f)})}{2(m^2-3m-4)}=\beta$

Or, $-(m^2+2)-\sqrt{\Delta(f)}< 2(m^2-3m-4) < -(m^2+2)+\sqrt{\Delta(f)}$

or, $-\sqrt{\Delta(f)} < m^2-6m-6 < \sqrt{\Delta(f)}$

And this inequality can be rewritten to:

$-\Delta(f) < (m^2-6m-6)^2 < \Delta(f)$

Now we're left with finding m such that $\Delta(f) > (m^2-6m-6)^2$

Thanks Dinky Doe i will give it a shot (Happy)
• Jun 27th 2010, 07:14 AM
Dinkydoe
Sorry, I made little error in writing out...

we must instead find m such that $\Delta(f) > (-m^2+6m+10)^2$

I hope you've noticed my error yourself, (see correction my previous post)

Notice Moreover that $\Delta(f) = (-m^2+6m+10)^2+12m^3-60m^2+24m+96$

good luck
• Jun 28th 2010, 03:11 AM
hooke
Quote:

Originally Posted by Dinkydoe
We have $\alpha,\beta = \frac{-(m^2+2)\pm \sqrt{m^4-44m^2+144m+196}}{2(m^2-3m-4)}$

Thus if we want to have $\alpha < -1 < \beta$ then we can impose 2 conditions:

1. $\Delta(f) = m^4-44m^2+144m+196> 0$

2. $\alpha =\frac{-(m^2+2) - \sqrt{\Delta(f)})}{2(m^2-3m-4)}< -1 <\frac{-(m^2+2) + \sqrt{\Delta(f)})}{2(m^2-3m-4)}=\beta$

How do you know that 2(m^2-3m-4) is positive?

Or, $-(m^2+2)-\sqrt{\Delta(f)}<- 2(m^2-3m-4) < -(m^2+2)+\sqrt{\Delta(f)}$

or, $-\sqrt{\Delta(f)} < -m^2+6m+10 < \sqrt{\Delta(f)}$

And this inequality can be rewritten to:

$-\Delta(f) < (-m^2+6m+10)^2 < \Delta(f)$

When you square the inequality here ie multiply by negative, shouldn't the inequality signs be flipped?

Now we're left with finding m such that $\Delta(f) > (-m^2+6m+10)^2$

Thanks Dinky doe, but i have a few questions ..
• Jun 28th 2010, 08:12 AM
Dinkydoe
Quote:

How do you know that 2(m^2-3m-4) is positive?
We don't. Not necessarily, but even it was negative, the inequality still holds. Both '<' signs flip and we can call $\beta = \alpha$, vice versa.

It doesn't quite matter what we call, $\alpha, \beta$

Quote:

When you square the inequality here ie multiply by negative, shouldn't the inequality signs be flipped?
I didn't actually square the inequality. But notice the following:

Let $a> 0$
If $-a < p < a$ holds then $|p|< a$ and this implies $-a^2
• Jun 29th 2010, 07:01 AM
hooke
Quote:

Originally Posted by Dinkydoe
We don't. Not necessarily, but even it was negative, the inequality still holds. Both '<' signs flip and we can call $\beta = \alpha$, vice versa.

It doesn't quite matter what we call, $\alpha, \beta$

I didn't actually square the inequality. But notice the following:

Let $a> 0$
If $-a < p < a$ holds then $|p|< a$ and this implies $-a^2

Can \alpha=\beta ?I thought its stated clearly that

\alpha<-1<\beta
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