The equation $x^4-6x^3-73x^2+kx+m=0$ has two positive roots, $\alpha$,
$\beta$ and two negative roots $\gamma$, $\delta$. It is given that
$\alpha\beta=\gamma\delta=4$. Show that
$\alpha^2-3(1+\sqrt{10})\alpha+4=0$, and find similar quadratic
equations satisfied by $\beta$, $\gamma$ and $\delta$.
Unsure how to approach this question. So far I have:
- Found $k=-24$ and $m=16$
- Shown $(\alpha+\beta)(\gamma+\delta)=-81$
- Found quadratic equation with roots $\alpha+\beta$ and $\gamma+\delta$ to be $x^2-6x-81=0$
- Found $\alpha+\beta=3+3\sqrt{10}$ and $\gamma+\delta=3-3\sqrt{10}$
Thanks for any help!
Best Answer
Almost there. You showed that $$\alpha+\beta=3+3\sqrt{10}$$ and you are given $$\alpha\beta=4$$ So a quadratic equation with roots $\alpha$ and $\beta$ is $$x^2-(\alpha+\beta)x+\alpha\beta=0$$ or $$x^2-3(1+\sqrt{10})x+4=0$$ Since you know $\alpha$ is root: $$\alpha^2-3(1+\sqrt{10})\alpha+4=0$$ You have the same equation for $\beta$. Apply similar reasoning for the other parameters.
Note you can multiply the above equation by $\beta$ and you have a linear equation for $\alpha$, in case you want to solve for $\alpha$ and you find quadratics difficult :)