Here is a problem from a timed exam:
If $\alpha$ and $\beta$ are roots of the equation $x^2-5x+2=0$, what is the value of $\dfrac{4\alpha+\beta^5}{5\beta^2}$?
$1) 21\qquad\qquad 2) 20\qquad\qquad3)19\qquad\qquad4)18$
Here is how I solved it:
We have $\alpha\beta=2$. Let's multiply the fraction by $\dfrac{\beta}{\beta}$:
$$\dfrac{4\alpha+\beta^5}{5\beta^2}=\dfrac{8+\beta^6}{5\beta^3}=\dfrac15\times\dfrac1{\beta^3}\times(8+\beta^6)$$ From $\alpha\beta=2$, we have $\dfrac1{\beta^3}=\dfrac{\alpha^3}8$, so the above expression is equal to: $$\dfrac1{40}\times(8\alpha^3+\alpha^3\beta^6)=\dfrac{1}{40}\times8(\alpha^3+\beta^3)=\dfrac15[(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)]=19$$
I'm looking for alternative approaches, especially the quicker ones, as it was taken from a timed exam.
