Question

alpha and ßeta are roots of x² + ax+b=0 and

k =(1-alpha) (1-beta)+(2-alpha) (2-beta)+(3-alpha) (3-beta) + ................................+(10-alpha) (10-beta).

If 1+2+3+....n=[n(n+1)]/2 and
1^2+2^2+3^3+.....+n^2=[n(n+1)(2n+1)]/6
Then k is equal to????

Answer 1

I'll use x and y to avoid all the alpha and beta stuff

(1-x)(1-y) = 1^2 -1(x+y) + xy
(2-x)(2-y) = 2^2 - 2(x+y) + xy
...
so the sum is
1^2 + 2^2 + ... + 10^2 - (1+2+...+10)(x+y) + 10xy
Now you can add it all up to find k