Question

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

• A. $(p^3+q)x^2–(p^3+2q)x+(p^3+q)=0$
• B. $(p^3+q)x^2–(p^3–2q)x+(p^3+q)=0$
• C. $(p^3-q)x^2–(5p^3–2q)x+(p^3–q)=0$
• D. $(p^3-q)x^2–(5p^3+2q)x+(p^3–q)=0$

Answer 1

The correct option is B

$(p^3+q)x^2–(p^3–2q)x+(p^3+q)=0$

Explanation for the correct option:

Formation of equation with roots:

Sum of roots $=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$​ and product$=1$

Given $\alpha +\beta =-p\text{,}{\alpha }^3+{\beta }^3=q,$

$$\begin{gathered} \Rightarrow (\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)=q\left[\because a^3+b^3=(a+b)(a^2+b^2-\mathrm{ab})\right] \\ \therefore {\alpha }^2+{\beta }^2-\alpha \beta =\frac{q}{-p}...\left(1\right) \\ {\mathrm{And}(\mathrm{\alpha }+\mathrm{\beta })}^2=p^2\mathbf{[}\mathbf{\because }{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{b}\mathbf{]} \\ {\alpha }^2+{\beta }^2+2\alpha \beta =p^2...\left(2\right) \end{gathered}$$

From $\left(1\right)$ and $\left(2\right)$, we get,

$$\begin{gathered} \left(2{\alpha }^2+2{\beta }^2-2\alpha \beta \right)+\left({\alpha }^2+{\beta }^2+2\alpha \beta \right)=\frac{2q}{-p}+p^2\left[\mathrm{Multiplying}2\mathrm{by}\left(1\right)\mathrm{and}\mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right)\right] \\ \Rightarrow 3{\alpha }^2+3{\beta }^2=\frac{p^3-2q}{p} \\ \Rightarrow 3\left({\alpha }^2+{\beta }^2\right)=\frac{p^3-2q}{p} \\ \Rightarrow \left({\alpha }^2+{\beta }^2\right)=\frac{p^3-2q}{3p} \end{gathered}$$

And

$$\begin{gathered} \left({\alpha }^2+{\beta }^2-\alpha \beta \right)-\left({\alpha }^2+{\beta }^2+2\alpha \beta \right)=\frac{q}{-p}-p^2\left[Subtracting\left(1\right)and\left(2\right)\right] \\ \Rightarrow -3\alpha \beta =\frac{p^3+q}{p} \\ \Rightarrow \alpha \beta =\frac{p^3+q}{3p} \end{gathered}$$

Therefore the required equation is

$$\begin{gathered} x^2-\frac{(p^3-2q)x}{(p^3+q)}+1=0 \\ \Rightarrow (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0 \end{gathered}$$

Hence, the correct option is (B)