Product and Sum of Polynomial Roots

Question

The ratio of the sum of the roots of the equation, $8x^3+px^2-2x+1=0 $ to the product of the roots of the equation $5x^3+7x^3-3x+q=0 $ is $3:2$. What is the value $\frac{p-q}{p+q}$?

Well I found out the sum of the roots of 1st equation is $-\frac{p}{8}$ and product of roots of second equation is $-\frac{d}{5}$.

Now what to do further? How to get in terms of $\frac{p-q}{p+q} $ ?

Answer 1

Using Vieta's Formulas,

the sum of the roots of $\ 8x^3+px^2-2x+1=0$ is $\frac{-p}8$

and the product of the roots of $\ 5x^3+7x^3-3x+q=0$ is $\frac {-q}5$

So, $$\frac {\frac{-p}8}{\frac {-q}5}=\frac32\implies \frac pq=\frac{3\cdot 8}{2\cdot 5}=\frac{12}5$$

Now, use Componendo and dividendo

or use $$\frac{p-q}{p+q}=\frac{\frac pq-1}{\frac pq+1}=\frac{\frac {12}5-1}{\frac {12}5+1}=\frac{12-5}{12+5}$$

or use $$\frac p{12}=\frac q5=r(\ne0)\text{(say)}\implies p=12r,q=5r$$

Answer 2

Hints:

The given ratio is (notice: is $\,q\,$ , not $\,d\,$ )

$$\frac{\frac p8}{\frac q5}=\frac{5p}{8q}$$

Get from here the solution to your question, whatever it is.