Polynomial help again ..

**mathaddict** · January 14th 2009, 11:14 PM

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 $\frac{pp'}{q}$ .

**NonCommAlg** · January 15th 2009, 12:01 AM

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.

**mathaddict** · January 15th 2009, 06:49 AM

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 ..

**Isomorphism** · January 15th 2009, 07:44 AM

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....

**mathaddict** · January 15th 2009, 11:16 PM

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 .

**Isomorphism** · January 15th 2009, 11:37 PM

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?

**mathaddict** · January 15th 2009, 11:50 PM

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 .

**Isomorphism** · January 15th 2009, 11:58 PM

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.

**mathaddict** · January 16th 2009, 12:03 AM

Thanks a lot iso , for clearing all my doubts ..

**Isomorphism** · January 16th 2009, 12:13 AM

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.

**Grandad** · January 16th 2009, 02:50 AM

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

**Isomorphism** · January 16th 2009, 08:53 AM

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?

**Grandad** · January 16th 2009, 10:04 AM

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

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