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$\alpha,\beta$ 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?

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2 Answers 2

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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 $\frac{m-5}3$.

$$1^5+\left(\frac{m-5}3\right)^5=33$$ $$\frac{m-5}3=2\Rightarrow m=11$$

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    $\begingroup$ +1 for mentioning "sum of the coefficients is 0 so one root is 1" because this is seldom included in common algebra text. $\endgroup$
    – Mick
    Jun 7, 2021 at 14:10
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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 $3x^2-9x+6=3(x-1)(x-2)=0$

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