Question

1/aA+b+1/aB+b where A and B are zeros of polynomial f(x)=ax²+bx+c

Answer 1

Answer:

Here α β are the roots of

ax²+bx+c=0

Therefore α+β=-b/a

αβ=c/a

Now α²/β+β²/α

=(α³+β³)/αβ

={(α+β)³-3αβ(α+β)}/αβ

={(-b/a)³ -3×(c/a)×(-b/a)}/(c/a)

={-b³/a³+3bc/a²}/(c/a)

=(3abc-b³/a³)×(a/c)

=(3abc-b³)/a²c

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