Question

\(\alpha\) and \(\beta\) are the zeroes of the polynomial \(5x^2 - 16x - 10\). Find the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\).

Hint:

Never waste time trying to solve the quadratic equation using the quadratic formula to find the decimal values of \(\alpha\) and \(\beta\).
For almost all symmetric expressions in \(\alpha\) and \(\beta\), you can express the equation cleanly using the sum and product formulas, which saves time and prevents calculation errors!

Answer 1

Step 1: Use the equation the roots satisfy, instead of the sum-of-squares identity.
Since $\alpha,\beta$ are roots of $5x^2-16x-10=0$, each satisfies $5x^2=16x+10$, so $\alpha^2=\frac{16\alpha+10}{5}$ and $\beta^2=\frac{16\beta+10}{5}$.
Step 2: Add these to get $\alpha^2+\beta^2$.
$\alpha^2+\beta^2=\frac{16(\alpha+\beta)+20}{5}$. Since $\alpha+\beta=-\frac{b}{a}=\frac{16}{5}$, this gives $\alpha^2+\beta^2=\frac{16\times\frac{16}{5}+20}{5}=\frac{\frac{256}{5}+\frac{100}{5}}{5}=\frac{356}{25}$.
Step 3: Divide by the product of the roots.
$\alpha\beta=\frac{c}{a}=\frac{-10}{5}=-2$, so $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{356/25}{-2}=-\frac{178}{25}$.
\[ \boxed{-\frac{178}{25}} \]