Question

If α and β are zeroes of the polynomial p(x)= x^2 + mx + n, then a polynomial whose zeroes are 1/apha , 1/beta is given by​

Answer 1

$\rm\Huge\underline{\text{Method 1}}$

$\rm\large\underline{\text{Substitution}}$

→ If the roots of the equation $\rm p(x)=0$ are $\rm x=\alpha,\beta$,  the roots of another equation are the reciprocals, $\rm x=\dfrac{1}{\alpha},\dfrac{1}{\beta}$.

$\rm\cdots\longrightarrow p(\dfrac{1}{x})=n+\dfrac{m}{x}+\dfrac{1}{x^2}$

$\rm\cdots\longrightarrow x^2p(\dfrac{1}{x})=nx^2+mx+1$

(As $\rm\dfrac{1}{\alpha},\dfrac{1}{\beta}$ are numbers, the first equation does not have $\rm x=0$ as a root, as well as the second equation.)

Hence $\rm nx^2+mx+1=0$ is the answer.

$\rm\Huge\underline{\text{Method 2}}$

$\rm\large\underline{\text{Relation between roots and coefficients}}$

→ Also called the Vieta's formula, the sum and product of roots in an equation is decided by,

$\rm\cdots\longrightarrow\red{\boxed{\text{(Sum)}=-\dfrac{\text{(Second-highest degree coefficient)}}{\text{(Leading coefficient)}}}}$

$\rm\cdots\longrightarrow\red{\boxed{\text{(Product)}=\dfrac{\text{(Constant term)}}{\text{(Leading coefficient)}}}}$

In a quadratic equation, $\rm ax^2+bx+c=0$, that is, for two roots $x=\alpha,\beta$,

$\rm\cdots\longrightarrow\red{\boxed{\alpha+\beta=-\dfrac{\text{b}}{\text{a}}}}$

$\rm\cdots\longrightarrow\red{\boxed{\alpha\beta=\dfrac{\text{c}}{\text{a}}}}$

Let's find the sum and product of roots.

$\rm\cdots\longrightarrow \alpha+\beta=-m$

$\rm\cdots\longrightarrow \alpha\beta=n$

Accordingly, the sum and the product of reciprocals are,

$\rm\cdots\longrightarrow \dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}=-\dfrac{m}{n}$

$\rm\cdots\longrightarrow \dfrac{1}{\alpha\beta}=\dfrac{1}{n}$

Consequently, the equation with two roots of above is,

$\rm\cdots\longrightarrow x^2+\dfrac{m}{n}x+\dfrac{1}{n}=0$

$\rm\cdots\longrightarrow nx^2+mx+1=0$

Hence $\rm nx^2+mx+1=0$ is the answer.