If px^2+qx+r=0, then sum of squares of roots it equal to....

[defeated_soldier]

If for the quadratic eqn px^2+qx+r=0, the sum of the squares of the roots is equal to sum of the cubes of the roots and q^3+pq^2=2p+3q=!0, then whats the value of pr ?

I want to solve the above problem. After reading the above problem , i noted down they key information as below:

. . .(alpha)^2 + (beta)^2 = (alpha)^3 + (beta)^3

(alpha,beta are roots of the quadratic eqn)[GIVEN]

From the quadratic eqn, I can write (alpha) + (beta) = -q/p

Damn!

Now what? I see nothing more. And I don't know why that "!=0" info has been given. I am completely lost.

defeated.

Kindly please don't give the solution, but please tell me how I should investigate this unknown problem. As you see, after writing those equations, my thinking power has been defeated, because I have no way out to solve this problem and I gave up finally.

Please tell me how you think you would solve this problem, what your thought process might be.

Thank you!

 

[defeated_soldier]

no response .

12 hour passed.

 

[Deleted member 4993]

Re: please help me to think

If for the quadratic eqn px^2+qx+r=0, the sum of the squares of the roots is equal to sum of the cubes of the roots and

q^3+pq^2=2p+3q=!0, ...This does not make sense

 

[morson]

sum of the squares of the roots of the first one: \(\displaystyle \frac{q^2 - 2pr}{p^2}\\)

sum of the cubes of the roots of the second cubic one can be found by some formula, i'm sure ... i just can't remember it ... all i know is that the sum of the cubes must be in the form \(\displaystyle \frac{q^2 - 2n}{p^2}\\). it would be laborious to use the cubic formula and cube all the general solutions..