Question

how to solve alpha4 and beta4

Answer 1

Let there be a quadratic polynomial ax^2+bx+c = 0 having 2 zeroes A and B. 
(A= Alpha and B= beta) 
so, A+B = -b/a and AB = c/a 
Now, 
A^4+B^4 = (A^2+B^2)^2 - 2A^2.B^2 
= [(A+B)^2 - 2AB]^2 - 2A^2.B^2 [A^2+B^2 = (A+B)^2 - 2AB] 
= [(A+B)^2 - 2AB]^2 - 2(AB)^2 
Now,A+B = -b/a and AB = c/a 

so, A^4+B^4 = [(A+B)^2 - 2AB]^2 - 2(AB)^2 = [(-b/a)^2 - 2c/a]^2 - 2(c/a)^2 
=(b^2/a^2 - 2c/a)^2 - 2c^2/a^2 
=(b^2/a^2)^2 +(2c/a)^2 - 2×(2c/a)×(b^2/a^2) - 2c^2/a^2 
= b^4/a^4 + 4c^2/a^2 - 4b^2c/a^3 - 2c^2/a^2 
=b^4/a^4 + 2c^2/a^2 - 4b^2c/a^3 
= (b^4 + 2c^2.a^2 - 4a.b^2.c)/a^4 

Hence, A^4+B^4 = (b^4 + 2c^2.a^2 - 4a.b^2.c)/a^4

Answer 2

Hey here is ur answer ....

For solving alpha 4 + beta 4 u can also write it as

(alpha sq.+ beta sq.)whole sq. - 2 alpha sq.beta sq.

hope it might help u .

any doubt comment