Question

how+to+solve+alpha4+and+beta4

Answer 1

Answer:

Step-by-step explanation:

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