Let p and q be real numbers such that p 0 , p 3 q and p 3 - q . If and are nonzero complex numbers satisfying = - p and 3 + 3 = q , then a quadratic equation having and as its roots is [2010]

Question

Let $p$ and $q$ be real numbers such that $p \neq 0$ , $p^{3} \neq q$ and $p^{3} \neq - q$ . If $α$ and $β$ are nonzero complex numbers satisfying $α + β = - p$ and $α^{3} + β^{3} = q$ , then a quadratic equation having $\frac{α}{β}$ and $\frac{β}{α}$ as its roots is [2010]

Smart Achievers · Accepted Answer

(2)

$Given : α + β = - p and α^{3} + β^{3} = q$

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

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

$Now for required quadratic equation,$

$Sum of roots = \frac{α}{β} + \frac{β}{α} = \frac{α^{2} + β^{2}}{α β}$

$= \frac{{(α + β)}^{2} - 2 α β}{α β}$ $= \frac{p^{2} - 2 (\frac{p^{3} + q}{3 p})}{\frac{p^{3} + q}{3 p}} = \frac{p^{3} - 2 q}{p^{3} + q}$

$and Product of roots = \frac{α}{β} \cdot \frac{β}{α} = 1$

$∴$ $Required equation is$ $x^{2} - (\frac{p^{3} - 2 q}{p^{3} + q}) x + 1 = 0$

$\Rightarrow (p^{3} + q) x^{2} - (p^{3} - 2 q) x + (p^{3} + q) = 0$

Solution

Q.
Let $p$ and $q$ be real numbers such that $p \neq 0$ , $p^{3} \neq q$ and $p^{3} \neq - q$ . If $α$ and $β$ are nonzero complex numbers satisfying $α + β = - p$ and $α^{3} + β^{3} = q$ , then a quadratic equation having $\frac{α}{β}$ and $\frac{β}{α}$ as its roots is [2010]

1 $(p^{3} + q) x^{2} - (p^{3} + 2 q) x + (p^{3} + q) = 0$

2 $(p^{3} + q) x^{2} - (p^{3} - 2 q) x + (p^{3} + q) = 0$

3 $(p^{3} - q) x^{2} - (5 p^{3} - 2 q) x + (p^{3} - q) = 0$

4 $(p^{3} - q) x^{2} - (5 p^{3} + 2 q) x + (p^{3} - q) = 0$

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Solution

Q. Let p and q be real numbers such that p≠0, p3≠q and p3≠-q. If α and β are nonzero complex numbers satisfying α+β=-p and α3+β3=q, then a quadratic equation having αβ and βα as its roots is [2010]

1 (p3+q)x2-(p3+2q)x+(p3+q)=0

2 (p3+q)x2-(p3-2q)x+(p3+q)=0

3 (p3-q)x2-(5p3-2q)x+(p3-q)=0