what is the formula of α^4+β^4=
Answers
Answered by
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
(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
ritikdubey17:
Thankyou very much
Answered by
1
hey mate
⬇⬇⬇⬇⬇
here is ur answer
⬇⬇⬇⬇⬇
a⁴+b⁴= ( a²-b²) (a²+b²) =(a-b) (a+b) ( a²+b²)
⬇⬇⬇⬇⬇
here is ur answer
⬇⬇⬇⬇⬇
a⁴+b⁴= ( a²-b²) (a²+b²) =(a-b) (a+b) ( a²+b²)
Similar questions