Question

Find a quadratic polynomial whose zeroes are $\dfrac{1}{\alpha }$ and $\dfrac{1}{\beta }$ . If $\alpha $ and $\beta $ are the zeroes of $2{{\text{x}}^2} - 3x + 5$.

Answer 1

Note: In this question we need to find a quadratic polynomial with given constraints. In order to find the polynomial we will use the formulae, $ - \dfrac{b}{a} = \alpha + \beta $ and $\dfrac{c}{a} = \alpha \beta $. Where, $\alpha $ and $\beta $ are the zeroes of $a{{\text{x}}^2} + bx + c = 0$. This will help us reach the answer.

Complete step-by-step answer:

We have been given the equation $2{{\text{x}}^2} - 3x + 5$ with $\alpha $ and $\beta $ as its zeroes.

So, $\dfrac{3}{2} = \alpha + \beta $ and $\dfrac{5}{2} = \alpha \beta $

Now, let the a quadratic polynomial whose zeroes are $\dfrac{1}{\alpha }$ and $\dfrac{1}{\beta }$ be $A{{\text{x}}^2} + Bx + C = 0$

So, $ - \dfrac{B}{A} = \dfrac{1}{\alpha } + \dfrac{1}{\beta }$ and $\dfrac{C}{A} = \dfrac{1}{{\alpha \beta }}$
$ \Rightarrow - \dfrac{B}{A} = \dfrac{{\beta + \alpha }}{{\alpha \beta }} = \dfrac{{\dfrac{3}{2}}}{{\dfrac{5}{2}}} = \dfrac{3}{5}$ and $\dfrac{C}{A} = \dfrac{1}{{\alpha \beta }} = \dfrac{1}{{\dfrac{5}{2}}} = \dfrac{2}{5}$

So, $ - \dfrac{B}{A} = \dfrac{3}{5}$ and $\dfrac{C}{A} = \dfrac{2}{5}$

Hence, A=5, B=-3 and C=2

Therefore, the required equation is $5{{\text{x}}^2} - 3x + 2 = 0$.

Note: In any quadratic polynomial:
A. The sum of the zeroes is equal to the negative of the coefficient of x by the coefficient of ${{\text{x}}^2}$.
B. The product of the zeroes is equal to the constant term by the coefficient of ${{\text{x}}^2}$.