# Polynomial help again ..

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• Jan 14th 2009, 11:14 PM
mathaddict
Polynomial help again ..
The roots of the quadratic equation $x^2+px+q=0$ , where q is not 0 are $\alpha$ and $\beta$ and the
roots of the quadratic equation $x^2+p'x+q=0$ is $k\alpha$ and $\frac{\beta}{k}$. For any p ,p' and q , show that the sum of the four possible values of k is $\frac{pp'}{q}$.
• Jan 15th 2009, 12:01 AM
NonCommAlg
Quote:

Originally Posted by mathaddict

The roots of the quadratic equation $x^2+px+q=0$ , where q is not 0 are $\alpha$ and $\beta$ and the
roots of the quadratic equation $x^2+p'x+q=0$ is $k\alpha$ and $\frac{\beta}{k}$. For any p ,p' and q , show that the sum of the four possible values of k is $\frac{pp'}{q}$.

so you have: $\begin{cases} \alpha + \beta=-p \\ \alpha \beta = q \\ k\alpha + \frac{\beta}{k}=-p' \end{cases}.$ fro the last equation you get $\alpha k^2 + \beta=-p'k,$ which with the first equation give us: $\alpha=\frac{p-p'k}{k^2-1}, \ \beta=\frac{(p'-pk)k}{k^2 - 1}.$ [here i'm assuming that $k \neq \pm 1.$]

putting what we got for $\alpha, \beta$ in the second equation, i.e. $\alpha \beta = q,$ gives us: $(p-p'k)(p'-pk)k=q(k^2-1)^2.$ thus: $qk^4 -pp'k^3 + (p^2 + p'^2 - 2q)k^2 - pp'k + q=0$ and the result follows.
• Jan 15th 2009, 06:49 AM
mathaddict
Re :
Quote:

Originally Posted by NonCommAlg
so you have: $\begin{cases} \alpha + \beta=-p \\ \alpha \beta = q \\ k\alpha + \frac{\beta}{k}=-p' \end{cases}.$ fro the last equation you get $\alpha k^2 + \beta=-p'k,$ which with the first equation give us: $\alpha=\frac{p-p'k}{k^2-1}, \ \beta=\frac{(p'-pk)k}{k^2 - 1}.$ [here i'm assuming that $k \neq \pm 1.$]

putting what we got for $\alpha, \beta$ in the second equation, i.e. $\alpha \beta = q,$ gives us: $(p-p'k)(p'-pk)k=q(k^2-1)^2.$ thus: $qk^4 -pp'k^3 + (p^2 + p'^2 - 2q)k^2 - pp'k + q=0$ and the result follows.

Thanks NonCommAlg , I understood what you did , but i don seem to get the result . The sum of all possible values of k ..
• Jan 15th 2009, 07:44 AM
Isomorphism
Quote:

Originally Posted by mathaddict
Thanks NonCommAlg , I understood what you did , but i don seem to get the result . The sum of all possible values of k ..

Hint: The four possible values are the roots of that quartic equation....
• Jan 15th 2009, 11:16 PM
mathaddict
RE :
This is what i did so far , trying to get the roots of this equation .

$qk^2-pp'k+(p^2+p'^2-2q)-\frac{pp'}{k}+\frac{q}{k^2}=0$

Let $w=k+\frac{1}{k^2}$ and substituting into the equation , i got

$q(w-2)-pp'w+p^2+p'^2-2q=0$

Till here , i have no idea how to get the roots .
• Jan 15th 2009, 11:37 PM
Isomorphism
You dont need to find the roots. All you need is the sum of the roots. Looking at the polynomial equation, cant you tell the sum of the roots in terms of its coefficients?
• Jan 15th 2009, 11:50 PM
mathaddict
Re :
Oh alright , my mistake . By the way , i understand for quadratic equation

$ax^2+bx+c=0$

The sum of roots is -b/a

But for quartic equation , i am not sure which is the sum of roots .
• Jan 15th 2009, 11:58 PM
Isomorphism
Quote:

Originally Posted by mathaddict
Oh alright , my mistake . By the way , i understand for quadratic equation

$ax^2+bx+c=0$

The sum of roots is -b/a

But for quartic equation , i am not sure which is the sum of roots .

Oh... sorry then :(

For a quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$, the sum of the roots is $-\frac{b}{a}$. Use this idea and you will immediately get the answer.
• Jan 16th 2009, 12:03 AM
mathaddict
Re :
Thanks a lot iso , for clearing all my doubts ..
• Jan 16th 2009, 12:13 AM
Isomorphism
I think it is a good thing to know Vieta's formulae to solve problems similar to the ones you have posed. You can read it from here.
• Jan 16th 2009, 02:50 AM
Grandad
Display problem
Quote:

Originally Posted by Isomorphism
I think it is a good thing to know Vieta's formulae to solve problems similar to the ones you have posed. You can read it from here.

I couldn't get this page to display correctly in my browser (Firefox). Any suggestions?

Grandad

• Jan 16th 2009, 08:53 AM
Isomorphism
Quote:

Originally Posted by Grandad
I couldn't get this page to display correctly in my browser (Firefox). Any suggestions?

Grandad

I use Firefox version 3.0.5 and it loads perfectly for me. Are you having problems with the TeX symbols?
• Jan 16th 2009, 10:04 AM
Grandad
Display problem
Quote:

Originally Posted by Isomorphism
I use Firefox version 3.0.5 and it loads perfectly for me. Are you having problems with the TeX symbols?

Just tried again - works fine now! (I also have Firefox 3.0.5.)

Grandad