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
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.
3 answers
idk... its wrong.. or not specific.. (abv one)
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
(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
2 answers