Question

if alpha,beta,gama are the zeroes of polynomial f(x)=x³–px²+qx–r then 1/alpha beta+1/betagama+1/gama alpha​

Answer 1

Answer:

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Step-by-step explanation:

α,β,γ zeroes of polynomial x

3

+px

2

+qx

2

+2 i.e, on keeping x=αorβorγ, we will get x

3

+px+qx+2=0

Also, αβ+1=0

We know that sum of roots of cubic polynomial =

1

−p

⇒α+β+γ=−p & αβ+βγ+γα=q

⇒αβγ=−2

Since αβγ=−2

& αβ+1=0⇒αβ=−1

⇒(−1)γ=−2

⇒γ=2

∴α+β+2=−p ⇒α+β=−p−2→(1)

αβ+βγ+γα=q⇒−1+2β+2α=q

⇒(α+β)=

2

(q+1)

→(2)

Equating (1)&(2)

⇒−p−2=

2

q+1

⇒−2p−4=q+1

⇒2p+q+5=0

Hence, the answer is 2p+q+5=0