If - and - are the zeros of polynomial x2-3 x-2- find 1-3-1-3A- 8-45B- 45-45C-D- 8-8

The correct option is C 45/8Given polynomial is : x^2+3x 2 ⇒α + β = b/a = 3/1 = 3 ⇒αβ = c/a = 2/1 = 2 ∴1/(α)^3 + 1/(β)^3 =(α)^3+(β)^3/(αβ)^3 =(α + ...

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Byju's Answer

Standard IX

Mathematics

Zeros of a polynomial

If α and β ar...

Question

If $α$ and $β$ are the zeros of polynomial
$x^{2} + 3 x - 2$ , find $\frac{1}{(α)^{3}} + \frac{1}{(β)^{3}}$ .

\frac{8}{45}

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\frac{- 8}{45}

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\frac{45}{8}

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\frac{- 45}{8}

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Solution

The correct option is C $\frac{45}{8}$
Given polynomial is :
$x^{2} + 3 x - 2$

$\Rightarrow α + β = \frac{- b}{a} = \frac{- 3}{1} = - 3$

$\Rightarrow α β = \frac{c}{a} = \frac{- 2}{1} = - 2$

$∴ \frac{1}{(α)^{3}} + \frac{1}{(β)^{3}}$

$= \frac{(α)^{3} + (β)^{3}}{(α β)^{3}}$

$= \frac{(α + β)^{3} - 3 α β (α + β)}{(α β)^{3}}$

$= \frac{(- 3)^{3} - 3 (- 2) (- 3)}{(- 2)^{3}}$

[Substituting the values of $α + β, α β$ ]

$= \frac{- 27 - 18}{- 8}$

$= \frac{45}{8}$

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If α and β are the zeros of polynomial x2+3x−2, find 1(α)3+1(β)3.

The correct option is C 458Given polynomial is : x2+3x−2 ⇒α+β=−ba=−31=−3 ⇒αβ=ca=−21=−2 ∴1(α)3+1(β)3 =(α)3+(β)3(αβ)3 =(α+β)3−3αβ(α+β)(αβ)3 =(−3)3−3(−2)(−3)(−2)3 [Substituting the values of α+β,αβ] =−27−18−8 =458

If $α$ and $β$ are the zeros of polynomial
$x^{2} + 3 x - 2$ , find $\frac{1}{(α)^{3}} + \frac{1}{(β)^{3}}$ .