If a,bare zeroes of the quadratic polynomial ax 2 +bx+c then find (1)a/b+b/a (2)a 2 +b 2
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α,β are the zeros of the quadratic polynomial ax²+bx+c=0.
Then, α+β=-b/a and α×β=c/a
(1) α/β+β/α
=(α²+β²)/αβ
={(α+β)²-2αβ}/αβ
={(-b/a)²-2c/a}/(c/a)
=(b²/a²-2c/a)/(c/a)
={(b²-2ac)/a²}/(c/a)
=a(b²-2ac)/a²c
=(b²-2ac)/ac
(2) α²+β²=(α+β)²-2αβ=(-b/a)²-2c/a=b²/a²-2c/a=(b²-2ac)/a²
Then, α+β=-b/a and α×β=c/a
(1) α/β+β/α
=(α²+β²)/αβ
={(α+β)²-2αβ}/αβ
={(-b/a)²-2c/a}/(c/a)
=(b²/a²-2c/a)/(c/a)
={(b²-2ac)/a²}/(c/a)
=a(b²-2ac)/a²c
=(b²-2ac)/ac
(2) α²+β²=(α+β)²-2αβ=(-b/a)²-2c/a=b²/a²-2c/a=(b²-2ac)/a²
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