If - - and - are the zeroes of the polynomial-fx-x3-px2-qx-r- then 1-1-1-

F(x)=x^3 px^2+qx r α,β and γ are zeroes of f(x) α+β+γ= ba= ( p)1=p αβ+βγ+γα=ca=q1=q αβγ= da= ( r)1=r 1αβ+1βγ+1αγ =γ+α+βαβγ=α+β+γαβγ =pr So, option d is correc ...

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Byju's Answer

Standard X

Mathematics

Relationship between Zeroes and Coefficients of a Polynomial

If α, β and γ...

Question

$I f α, β a n d γ are the zeroes of the polynomial, f (x) = x^{3} - p x^{2} + q x - r, t h e n \frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α} =$

- p / r

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r / p

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- r / p

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p / r

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Solution

The correct option is D
$p / r$

$f (x) = x^{3} - p x^{2} + q x - r α, β a n d γ a r e z e r o e s o f f (x) α + β + γ = \frac{- b}{a} = \frac{- (- p)}{1} = p α β + β γ + γ α = \frac{c}{a} = \frac{q}{1} = q α β γ = \frac{- d}{a} = \frac{- (- r)}{1} = r \frac{1}{α β} + \frac{1}{β γ} + \frac{1}{α γ} = \frac{γ + α + β}{α β γ} = \frac{α + β + γ}{α β γ} = \frac{p}{r}$
So, option d is correct.

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If α, β and γ are the zeroes of the polynomial,f(x)=x3−px2+qx−r, then 1αβ+1βγ+1γα=

The correct option is D p/r f(x)=x3−px2+qx−rα,β and γ are zeroes of f(x)α+β+γ=−ba=−(−p)1=pαβ+βγ+γα=ca=q1=qαβγ=−da=−(−r)1=r1αβ+1βγ+1αγ=γ+α+βαβγ=α+β+γαβγ=pr So, option d is correct.

$I f α, β a n d γ are the zeroes of the polynomial, f (x) = x^{3} - p x^{2} + q x - r, t h e n \frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α} =$