$\alpha,\beta$ are roots of the equation $3x^2-(m-2)x+(m-5)=0$ such that $\alpha^5+\beta^5=33$. Find the value of $m$.
$$\alpha+\beta=\frac{m-2}3$$
Squaring and cubing it one by one and then multiplying them, and putting $$\alpha^5+\beta^5=33\\\alpha\beta=\frac{m-5}3,$$ I got a quintic equation in $m$ which I couldn't solve. Any help?
Also, any other method to approach this question?
Best Answer
We have, $3x^2-(m-2)x+(m-5)=0$. Notice that sum of the coefficients is $0$, so one root is $1$ and another root is $\frac{m-5}3$.
$$1^5+\left(\frac{m-5}3\right)^5=33$$ $$\frac{m-5}3=2\Rightarrow m=11$$