# Polynomial help again ..

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

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

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

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

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

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

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
RE :
This is what i did so far , trying to get the roots of this equation .

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

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

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

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
Re :
Oh alright , my mistake . By the way , i understand for quadratic equation

$\displaystyle ax^2+bx+c=0$

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:

Oh alright , my mistake . By the way , i understand for quadratic equation

$\displaystyle ax^2+bx+c=0$

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 $\displaystyle ax^4 + bx^3 + cx^2 + dx + e = 0$, the sum of the roots is $\displaystyle -\frac{b}{a}$. Use this idea and you will immediately get the answer.
• Jan 16th 2009, 12:03 AM
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
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?

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

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

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