Roots of a Polynomial Question

Question

If $\alpha$ and $\beta$ are roots of the equation $x^2+mx+n=0$ , find the roots of $nx^2+(2n-m^2)x+n=0$ in terms of $\alpha$ and $\beta$.

I really need help with this problem. I've started by finding the sum and products of roots of the first equation and don't know what to do next.

Hint $ $ Its root product $=1$ and root sum $= m\left[\dfrac{m}n\right]-2, =, (\alpha+\beta)\left[\dfrac{1}\alpha+\dfrac{1}\beta\right] -2, =, \dfrac{\alpha}\beta+\dfrac{\beta}\alpha$ $\qquad\qquad\qquad $ — Bill Dubuque, Jan 22 '19 at 18:26

score 5 · Answer 1 · answered Jan 22 '19 at 11:15

5

Substituting $m,n$ obtained from Vieta, the second equation is

$$\alpha\beta x^2+(2\alpha\beta-(\alpha+\beta)^2)x+\alpha\beta=0$$

or

$$x^2-\left(\frac\alpha\beta+\frac\beta\alpha\right)x+1=0.$$

Again by Vieta, the roots are obvious.

answered Jan 22 '19 at 11:15

score 1 · Answer 2 · answered Jan 22 '19 at 11:11

1

Hint: The discriminant of $nx^2+(2n-m^2)x+n$ is $m^4 - 4 m^2 n=m^2(m^2-4n)=(\alpha+\beta)^2(\alpha-\beta)^2$.

answered Jan 22 '19 at 11:11

lhf

score 0 · Answer 3 · answered Jan 22 '19 at 11:08

0

Try writing $m$ and $n$ in terms of $\alpha$ and $\beta$. Then solve the second equation in the usual way. Hint: $x^2+mx+n = (x-\alpha)(x-\beta)$.

answered Jan 22 '19 at 11:08

Klaus

3 Answers3