Finding all possible values of $d$ given that the sum of the reciprocals of a polynomial is $17$

Question

Let $p(x) = x^5 - 833x^4 + ax^3 + bx^2 + cx + d$ such that the roots of $p(x)$ are in geometric progression. If the sum of the reciprocal of the roots is $17,$ determine all possible values of $d.$

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I was thinking of trying to use Vieta's Formulas, but I'm not sure how to apply them. Could someone give me a hint please?

Answer 1

Hint: Vieta's Formulas is the way to go.

Let the roots be $a,a q,a q^2,a q^3,a q^4$. Then $a+a q+a q^2+a q^3+a q^4=833$. Use this to simplify $$ \frac{1}{a}+\frac{1}{a q}+\frac{1}{a q^2}+\frac{1}{a q^3}+\frac{1}{a q^4} = 17 $$ and $$ d = -a (a q) (a q^2) (a q^3) (a q^4) $$

Answer 2

Take roots as $\frac s{r^2}, \frac s{r}, s, sr, sr^2$. Using Vieta's formula, $$\frac s{r^2}+ \frac s{r}+ s+ sr+ sr^2=833$$ It's a GP with first term $\frac s{r^2}$ and common ratio $r$. So, we get $1st$ equation: $$\frac{\frac{s}{r^2}(r^5-1)}{r-1}=833$$ Also, given that the sum of reciprocals $=17$. So, $$\frac {r^2}s+ \frac rs+ \frac1s+ \frac1{sr}+ \frac1{sr^2}=17$$ It's a GP too with first term $\frac{r^2}s$ and common ratio $\frac1r$. So, we get $2nd$ equation: $$\frac{\frac{r^2}{s}({1-(\frac1r)^5)}}{1-\frac1r}=17$$ $$\implies \frac{r^2 (r^5-1)}{sr^5(\frac{r-1}r)}=17$$ $$\implies\frac{r^5-1}{sr^2(r-1)}=17$$ Dividing $1st$ equation by $2nd$, we get, $$s^2=49$$ $$\implies s=\pm7$$ Now, again using Vieta's formula, $$\frac s{r^2}\cdot\frac sr\cdot s\cdot sr\cdot sr^2=-d$$ $$\implies s^5=-d$$ So, the possible values of $d$ are $\pm7^5$.