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

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.

3. Re :

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 ..

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....

5. 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 .

6. 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?

7. 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 .

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.

9. Re :

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

10. 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.

11. Display problem

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.
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