Question

If α and β are two zeroes of the polynomial f(x) = ax^2 + by + c, then evaulate α^4+ β^4 .

Answer 1

hi friend,

actually ur question should be ax²+bx+c but not ax²+by+c

given α and β are two zeroes of the polynomial f(x) = ax^2 + bx + c

then,œ+ß=-b/a

ϧ=c/a

»» now, œ⁴+ß⁴ = (œ²+ß²)²-2œ²ß²

=((œ+ß)²-2œß)²-2œ²ß²

(b²/a²-2c/a)²-2(c²/a²)

=( b²-2ac)²/a⁴ -2c²/a²

=b⁴+4a²c²-4b²ac-2a²c²/a⁴

=b⁴+2a²c²-4b²ac/a⁴

I hope this will help u :)

Answer 2

Given f(x) = ax^2 + bx + c (I'm assuming you meant 'bx' and not 'by')

α and ß are zeroes of the equation. We know, α+ß = -b/a and αß = c/a A/Q

α^4 + ß^4 = (α^2 + b^2)^2 - 2(α^2)(ß^2)....(1)

Let's find α^2+ß^2 (You can do it direct. I'm separating just for the sake of understanding and making it seem clear)

α^2+ß^2 = (α+ß)^2 - 2^αß
= (-b/a)^2 - 2(c/a)
= b^2/a^2 - 2c/a
= (b^2-2ca)/a^2....(2)

Putting (2) value in (1)

α^4 + ß^4 = (α^2 + b^2)^2 - 2(α^2)(ß^2)
= { (b^2-2ca)/a^2 }^2 - 2(c/a)^2
= { (b^2-2ca)^2/ a^4 } - 2c^2/a^2
= (b^4 + 4c^2a^2- 4ab^2c)/a^4 - 2c^2/a^2

So,

α^4 + ß^4 = (b^4 + 4c^2a^2- 4ab^2c)/a^4 - 2c^2/a^2