Question

Let p and q be real numbers such that \[p \ne 0,{p^3} \ne q\] and ${P^3} \ne - q.$ If α and β are non-zero complex numbers satisfying \[{\alpha + \beta = - p}\] and \[{{\alpha }^{3}} + {{\beta }^{3}}{ = q}\] then a quadratic equation having \[\dfrac{\alpha }{\beta } and \dfrac{\beta }{\alpha }\] as its roots is
A. $\left( {{{\text{p}}^{\text{3}}}{\text{ + q}}} \right){{\text{x}}^{\text{2}}}{\text{ - (}}{{\text{p}}^{\text{3}}}{\text{ + 2q)x + (}}{{\text{p}}^{\text{3}}}{\text{ + q) = 0}}$
B. $\left( {{{\text{p}}^{\text{3}}}{\text{ + q}}} \right){{\text{x}}^{\text{2}}}{\text{ - (}}{{\text{p}}^{\text{3}}}{\text{ - 2q)x + (}}{{\text{p}}^{\text{3}}}{\text{ + q) = 0}}$
C. $\left( {{{\text{p}}^{\text{3}}}{\text{ - q}}} \right){{\text{x}}^{\text{2}}}{\text{ - (5}}{{\text{p}}^{\text{3}}}{\text{ - 2q)x + (}}{{\text{p}}^{\text{3}}}{\text{ - q) = 0}}$
D. $\left( {{{\text{p}}^{\text{3}}}{\text{ - q}}} \right){{\text{x}}^{\text{2}}}{\text{ - (5}}{{\text{p}}^{\text{3}}}{\text{ + 2q)x + (}}{{\text{p}}^{\text{3}}}{\text{ + q) = 0}}$

Answer 1

Hint: To solve this question, we need to know the basic theory related to the quadratic equation. Here first we will assume that the new equation be ${x^2} + Bx + C = 0$and then after calculate the value of B and C. So that we will get a quadratic equation having \[\dfrac{\alpha }{\beta } and \dfrac{\beta }{\alpha }\] as its roots.

Complete step-by-step answer:
Let the new equation be
${x^2} + Bx + C = 0$
${\text{B = }}$\[ - \left( {\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha }} \right)\]...........Sum of the roots =$ - \dfrac{b}{a}$
C=\[\dfrac{\alpha}{\beta} \cdot \dfrac{\beta }{\alpha }\]................Product of roots =$ - \dfrac{c}{a}$
$ \Rightarrow $\[{{\text{x}}^{\text{2}}}{\text{ - }}\left( {\dfrac{{\alpha }}{{\beta
}}{\text{ + }}\dfrac{{\beta }}{{\alpha }}} \right){\text{x + }}\dfrac{{\alpha
}}{{\beta }}{\times }\dfrac{{\beta }}{{\alpha }}{\text{ = 0}}\]
$ \Rightarrow $ ${{\text{x}}^{\text{2}}}{\text{ - }}\dfrac{{{\text{(}}{{\alpha}^{\text{2}}}{\text{ + }}{{\beta}^{\text{2}}}{\text{)}}}}{{{\alpha \beta}}}{\text{x + 1 = 0}}$
$ \Rightarrow $ ${{\text{x}}^{\text{2}}}{\text{ - }}\dfrac{{{{{\text{(}}{{\alpha}^{}}{\text{ + }}{{\beta}^{}}{\text{)}}}^2} - 2{\alpha \beta}}}{{{\alpha \beta}}}{\text{x + 1 = 0}}$
Now, we have ${\alpha ^3} + {\beta ^3} = q$
$ \Rightarrow $ ${{\text{(}}{{\alpha}^{}}{\text{ + }}{{\beta}^{}}{\text{)}}^3}$-3${\alpha \beta}$${(\alpha + \beta) = q}$
$ \Rightarrow $ ${\text{ - }}{{\text{p}}^{\text{3}}}{ + 3p\alpha \beta = q}$
$ \Rightarrow $ ${\alpha \beta = }\dfrac{{{\text{q + }}{{\text{p}}^{\text{3}}}}}{{{\text{3p}}}}$
$ \Rightarrow $ ${{\text{x}}^{\text{2}}}{\text{ - }}\dfrac{{{{\text{p}}^{\text{2}}}{\text{ - 2}}\left( {\dfrac{{{{\text{p}}^{\text{3}}}{\text{ + q}}}}{{{\text{3p}}}}} \right)}}{{\dfrac{{{{\text{p}}^{\text{3}}}{\text{ + q}}}}{{{\text{3p}}}}}}{\text{x + 1 = 0}}$
$ \Rightarrow $ p$\left( {{{\text{p}}^{\text{3}}}{\text{ + q}}} \right){{\text{x}}^{\text{2}}}{\text{ - (3}}{{\text{p}}^{\text{3}}}{\text{ - 2}}{{\text{p}}^{\text{3}}}{\text{ - 2q)x + (}}{{\text{p}}^{\text{3}}}{\text{ + q) = 0}}$
$ \Rightarrow $ $\left( {{{\text{p}}^{\text{3}}}{\text{ + q}}} \right){{\text{x}}^{\text{2}}}{\text{ - (}}{{\text{p}}^{\text{3}}}{\text{ - 2q)x + (}}{{\text{p}}^{\text{3}}}{\text{ + q) = 0}}$
Therefore, a quadratic equation having \[\dfrac{\alpha }{\beta }and\dfrac{\beta }{\alpha }\] as its roots is $\left( {{{\text{p}}^{\text{3}}}{\text{ + q}}} \right){{\text{x}}^{\text{2}}}{\text{ - (}}{{\text{p}}^{\text{3}}}{\text{ - 2q)x + (}}{{\text{p}}^{\text{3}}}{\text{ + q) = 0}}$.
Thus, option (B) is the correct answer.

Note: Always remember that a quadratic polynomial, when equated to zero, becomes a quadratic equation. The values of x satisfying the equation are called the roots of the quadratic equation. And also, A quadratic equation becomes an identity (a, b, c = 0) if the equation is satisfied by more than two numbers i.e. having more than two roots or solutions either real or complex.