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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 ...

Let p and q be real numbers such that p≠0,p3≠qp \ne 0,{p^3} \ne q and P3≠−q.{P^3} \ne - q. If α and β are non-zero complex numbers satisfying α+β=−p{\alpha + \beta = - p} and α3+β3=q{{\alpha }^{3}} + {{\beta }^{3}}{ = q} then a quadratic equation having αβandβα\dfrac{\alpha }{\beta } and \dfrac{\beta }{\alpha } as its roots is
A. (p3 + q)x2 - (p3 + 2q)x + (p3 + q) = 0\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. (p3 + q)x2 - (p3 - 2q)x + (p3 + q) = 0\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. (p3 - q)x2 - (5p3 - 2q)x + (p3 - q) = 0\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. (p3 - q)x2 - (5p3 + 2q)x + (p3 + q) = 0\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}}

Explanation

Solution

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 x2+Bx+C=0{x^2} + Bx + C = 0and then after calculate the value of B and C. So that we will get a quadratic equation having αβandβα\dfrac{\alpha }{\beta } and \dfrac{\beta }{\alpha } as its roots.

Complete step-by-step answer :
Let the new equation be
x2+Bx+C=0{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 (p3 + q)x2 - (p3 - 2q)x + (p3 + q) = 0\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.