Math
posted by Steve
If alpha and beta are the zeros
of the polynomial ax^2 + bx + c
then evaluateA. (alpha)^2 / beta +
(beta)^2 / alpha
B. alpha^2 .beta + alpha.beta^2
C. 1/(alpha)^4 + 1/(beta)^4.
Please work the complete solution.

Reiny
let the two roots be m and n
then we want
m^2/n + n^2/m
= (m^3 + n^3)/(mn)
= (m+n)(m^2 mn + n^2)/(mn) , where m^2 + n^2 = (m+n)^2  2mn
= (m+n)( (m+n)^2  3mn)/(mn)
now from a^2 + bx + c = 0
m+n = b/a
and mn = c/a
(m+n)( (m+n)^2  3mn)/(mn)
= (b/a)( (b/a)^2  3(c/a) )/(c/a)
which I reduced to
b^3/(ca^2)  3
check my algebra and do the others the same way 
dhami
idk... its wrong.. or not specific.. (abv one)

Prahlad
We know for any quadratic polynomial f(x)=ax^2+bx+c with roots alpha(p) and beta(q)
(xp)(xq)= K[x^2(p+q)x+pq]
So we express (p+q) as b/a and pq as c/a.....
A.)(p^2/q) + (q^2/p)=?
By simply taking LCM, we can write the above statement as
(p^3+q^3)/pq
=(p+q)(p^2+q^2pq)[identity used]
{Now what you must understand here is that we can only substitute the values of the sum and products of the roots so our attempt now must be towards expressing this in the form of (p+q) or pq only}
=(p+q)((p+q)^23pq)
On reducing by substitution
You will obtain (3abcb^3)/a^3
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