# $\alpha,\beta$ are roots of the equation $3x^2-(m-2)x+(m-5)=0$ such that $\alpha^5+\beta^5=33$. Find the value of $m$.

$$\alpha,\beta$$ 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?

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 $$\frac{m-5}3$$.
$$1^5+\left(\frac{m-5}3\right)^5=33$$ $$\frac{m-5}3=2\Rightarrow m=11$$
Observe, $$\alpha+\beta=\frac{m-2}{3}\;\;\;\text{and}\;\;\; \alpha\beta=\frac{m-5}{3}\implies \alpha+\beta=\alpha\beta+1\implies (\alpha-1)(\beta-1)=0$$ Therefore, either $$\alpha=1$$ or $$\beta=1$$. Since $$\alpha^5+\beta^5=33,$$ the roots of the quadratic must be $$1$$ and $$2$$. Plugging this into one of the previous equations, we must have $$\boxed{m=11}$$
Note: The quadratic equation is $$3x^2-9x+6=3(x-1)(x-2)=0$$