Question

if alpha and bita are zeroes of the quadratic polynomial ax2+bx+c,then find a quadratic polynomial whose zeroes are

1/alpha and 1/bita.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{polynomial}\mathrm{is}, \\ p\left(x\right)=a{x}^2+bx+c \\ \mathrm{Since}\alpha \&\beta \mathrm{are}\mathrm{zeroes}\mathrm{of}p\left(x\right),\mathrm{then} \\ \alpha +\beta =-\frac{b}{a} \\ \alpha \times \beta =\frac{c}{a} \\ \mathrm{Now},\frac{1}{\alpha }\mathrm{and}\frac{1}{\beta }\mathrm{are}\mathrm{the}\mathrm{zeroes}\mathrm{of}\mathrm{required}\mathrm{polynomial} \\ \mathrm{Sum}\mathrm{of}\mathrm{zeroes}\mathrm{of}\mathrm{required}\mathrm{polynomial},\mathrm{S}=\frac{1}{\alpha }+\frac{1}{\beta } \\ =\frac{\beta +\alpha }{\alpha \beta } \\ =\frac{-{\displaystyle \raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}}{{\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}}=-\frac{b}{c} \\ \mathrm{Product}\mathrm{of}\mathrm{zeroes}\mathrm{of}\mathrm{required}\mathrm{polynomial},\mathrm{P}=\frac{1}{\alpha }\times \frac{1}{\beta } \\ =\frac{1}{\alpha \beta }=\frac{a}{c} \end{gathered}$$

$$\begin{gathered} \mathrm{Hence},\mathrm{required}\mathrm{polynomial}\mathrm{is}, \\ f\left(x\right)=k\left(x^2-\mathrm{S}x+\mathrm{P}\right),k\mathrm{is}\mathrm{non}\mathrm{zero}\mathrm{real}\mathrm{number} \\ =k\left[x^2+\frac{b}{c}x+\frac{a}{c}\right] \\ =\frac{k}{c}\left[c{x}^2+bx+a\right] \\ \mathrm{Taking}k=c,\mathrm{we}\mathrm{get}, \\ f\left(x\right)=c{x}^2+bx+a \end{gathered}$$