Finding sum of zeroes of $f(x)$ from $g'(x)=60f(x)$ .

Question

Question:

Let $f(x)=x^{5}+a x^{4}+b x^{3}+c x^{2}+d x+e$. If $ad-25e=0$ and $g(x)=10 x^{6}+12 a x^{5}+15 b x^{4}+20 c x^{3}+30 d x^{2}+60 e x+k$ where $a, b, c, d, e, k \in R $ has six positive real zeros and $f(3)=1$, then find the sum of the zeros of $f(x)$.

Things I have observed: $$g(x)=10 x^{6}+12 a x^{5}+15 b x^{4}+20 c x^{3}+30 d x^{2}+60 e x+k\\g'(x)=60(x^{5}+a x^{4}+b x^{3}+c x^{2}+d x+e)\\g'(x)=60f(x)$$ $$g'(3)=60f(3)=60$$

Also by Descartes' rule of signs, the sign of $a,b,c,d,e$ must be alternate.

Further, I tried using Vieta's formula but it didn't lead me anywhere. I have not been able to extract information from $ad=25e$.

Can someone please give me hints? (Please give me hints only, it helps me!)

Thanks

Answer 1

Hints:

1. Using Vieta's relations, show $ad=25e \implies \displaystyle \left(\sum_i r_i\right)\cdot \left(\prod_i r_i\right)\left(\sum_i \frac1{r_i} \right)=25\left(\prod_i r_i\right)$, where $r_i$ denote the roots of $f$.
2. Cancel the non-zero (why?) product on both sides and compare to what CS inequality (why does that apply here?) would say. You should be able to write the form of $f$ now in only one unknown parameter.
3. Now use the information $f(3)=1$ to write out $f$ and hence the roots or their sum.