Proving roots of quadratic equations

Question

While working out, for the required quadratic equation, my result is: $x^2-\left(\frac{{\beta }^3+{\alpha }^3}{{(\alpha \beta )}^3}\right)x+\frac{1}{{\left(\alpha \beta \right)}^3}$

I am unable to move to the next part of the question. Any help will be appreciated.

Answer 1

• If $\alpha$, $\beta$ are the roots of $ax^2+bx+c$ and $\alpha\beta^2=1$, prove that $a^3+c^3+abc=0$.

\begin{align*} \alpha\beta^2&=1\\ \frac ca\beta&=1\\ \frac ca\left(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\right)&=1\\ -bc\pm c\sqrt{b^2-4ac}&=2a^2\\ \pm c\sqrt{b^2-4ac}&=2a^2+bc\\ \left(\pm c\sqrt{b^2-4ac}\right)^2&=(2a^2+bc)^2\\ b^2c^2-4ac^3&=4a^4+b^2c^2+4a^2bc\\ 4a^4+4a^2bc+4ac^3&=0\\ a^3+abc+c^3&=0\\ \end{align*}

Answer 2

so $\alpha = \frac {-b \pm\sqrt{b^2 - 4ac}}{2a}$ and let $\beta =\frac {-b \mp\sqrt{b^2 - 4ac}}{2a}$

so $\alpha\beta=\frac {-b +\sqrt{b^2 - 4ac}}{2a}\frac {-b -\sqrt{b^2 - 4ac}}{2a}=$

$\frac {b^2 - \sqrt{b^2 - 4ac}^2}{4a^2} = \frac {4ac}{4a^2} = \frac ca$.

And so if $\alpha\beta^2 =1$ then $\beta = \frac 1{\alpha\beta} = \frac ac$

Plug $x = \beta = \frac ac$ into $ax^2 + bx + c = 0$

and we get $a\beta^2 + b\beta + c = $

$\frac {a^3}{c^2}+\frac {ab}c + c= 0$

so $ a^3+abc + c^3 = 0$