Question

If $\alpha ,\beta$ are the zeroes of the polynomial $x^2+bx+c$, then polynomial having $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ then its zeroes is

• A. $x^2+cx+b$
• B. $b{x}^2+cx+a$
• C. $c{x}^2+bx+1$
• D. $x^2-bx+c$

Answer 1

The correct option is C $c{x}^2+bx+1$

Since $\alpha ,\beta$ are the zeros of the polynomial $x^2+bx+c$
$\Rightarrow \alpha +\beta =-b$
$\alpha \beta =c$
$\text{we have to find the polynomial whose zeroes are}$
$\frac{1}{\alpha }$ and $\frac{1}{\beta }$
$\text{Which is}$
$x^2-Sx+P$
where $S=\frac{1}{\alpha }+\frac{1}{\beta }$
$=\frac{\alpha +\beta }{\alpha \beta }$
$\therefore S=\frac{-b}{c}$
Now, $P=\frac{1}{\alpha }\times \frac{1}{\beta }$
$=\frac{1}{\alpha \beta }$
$\therefore P=\frac{1}{c}$
$\therefore x^2-Sx+P=x^2-\frac{-b}{c}x+\frac{1}{c}$
$=c{x}^2+bx+1$
Option B is correct.