Let $p$ and $q$ be real numbers such that $p
eq 0,\ p^{3}
eq q$ and $p^{3}
eq - q$. If $\alpha$ and $\beta$ are non-zero complex numbers satisfying $\alpha + \beta = - p$ and $\alpha^{3} + \beta^{3} = q$, then a quadratic equation having $\frac{\alpha}{\beta}\$and$\ \frac{\beta}{\alpha}$ as its roots is

Question

Let $p$ and $q$ be real numbers such that $p 
eq 0,\ p^{3} 
eq q$ and $p^{3} 
eq - q$. If $\alpha$ and $\beta$ are non-zero complex numbers satisfying $\alpha + \beta = - p$ and $\alpha^{3} + \beta^{3} = q$, then a quadratic equation having $\frac{\alpha}{\beta}\$and$\ \frac{\beta}{\alpha}$ as its roots is

Dr. Abubacker · Accepted Answer

(b)
Sum of roots$= \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}$ and product =1
Given, $\alpha + \beta = - p$ and $\alpha^{3} + \beta^{3} = q$
$\Rightarrow (\alpha + \beta)\left( \alpha^{2} - \alpha\beta + \beta^{2} ight) = q$
$\ 	herefore\ \alpha^{2} + \beta^{2} - \alpha\beta = \frac{- q}{p}$ ...(i)
And $(\alpha + \beta)^{2} = p^{2}$
$\Rightarrow \ \alpha^{2} + \beta^{2} + 2\alpha\beta = p^{2}$
From Eqs. (i) and (ii), we get
$\alpha^{2} + \beta^{2} = \frac{p^{3} - 2q}{3p}$
And $\alpha\beta = \frac{p^{3} + q}{3p}$
$	herefore$ Required equation is
$x^{2} - \frac{(p^{3} - 2q)x}{(p^{3} + q)} + 1 = 0$
$\Rightarrow \left( p^{3} + q ight)x^{2} - \left( p^{3} - 2q ight)x + \left( p^{3} + q ight) = 0$

Explanation

Reviewed by Dr. Abubacker

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