# Math

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.

1. 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

posted by Reiny
2. idk... its wrong.. or not specific.. (abv one)

posted by dhami
3. We know for any quadratic polynomial f(x)=ax^2+bx+c with roots alpha(p) and beta(q)
(x-p)(x-q)= 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^2-pq)[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)^2-3pq)
On reducing by substitution-
You will obtain (3abc-b^3)/a^3

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