If - - and - - are non-zero complex members satisfying - - - - - - p- and - - 3 - - 3- - q then the quadratic equation having - and - as it roots is -

α + β = p , α^3 + β^2 = q α^3 + β^3 = (α + β)^3 3αβ(α + β) = ( p)^3 3α p( p) = q αβ = q + p^33p #160; (I) α + β = p ⇒ (α + β)^2 = p^ ...

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Standard XII

Mathematics

Complex Numbers

If 'α ' and...

Question

If $^{'} α^{'}$ and $^{'} β^{'}$ are non-zero complex members satisfying $[α + β = - p]$ and $[α^{3} + β^{3}] = q$ then the quadratic equation having $\frac{α}{β}$ and $\frac{β}{α}$ as it roots is ______.

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Solution

$α + β = p, α^{3} + β^{2} = q$
$α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β) = (- p)^{3} - 3 α p (- p) = q$
$α β = \frac{q + p^{3}}{3 p} - (I)$
$α + β = - p \Rightarrow (α + β)^{2} = p^{2} \Rightarrow α^{2} + β^{2} + 2 α β = p^{2}$
$α^{2} + β^{2} = p^{2} - 2 α β = \frac{p^{3} - 2 q}{3 p} - (I I)$
$\frac{α}{β} + \frac{β}{α} - \frac{α^{2} + β^{2}}{α β} = \frac{p^{3} - 2 q}{p^{3} + q}$
Equation will be
$\to x^{2} - (\frac{α}{β} + \frac{β}{α}) x + (\frac{α}{β} . \frac{β}{α})$

$\to x^{2} - (\frac{p^{3} - 2 q}{p^{3} + q}) x + 1$

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If ′α′ and ′β′ are non-zero complex members satisfying [α+β=−p] and [α3+β3]=q then the quadratic equation having αβ and βα as it roots is ______.

α+β=p,α3+β2=qα3+β3=(α+β)3−3αβ(α+β)=(−p)3−3αp(−p)=qαβ=q+p33p−(I)α+β=−p⇒(α+β)2=p2⇒α2+β2+2αβ=p2α2+β2=p2−2αβ=p3−2q3p−(II)αβ+βα−α2+β2αβ=p3−2qp3+qEquation will be→x2−(αβ+βα)x+(αβ.βα)→x2−(p3−2qp3+q)x+1

If $^{'} α^{'}$ and $^{'} β^{'}$ are non-zero complex members satisfying $[α + β = - p]$ and $[α^{3} + β^{3}] = q$ then the quadratic equation having $\frac{α}{β}$ and $\frac{β}{α}$ as it roots is ______.