If $\mathrm{\alpha }$ and $\mathrm{\beta }$ are the roots of the equation ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ and if ${\mathrm{px}}^2+\mathrm{qx}+\mathrm{r}=0$ has roots $\frac{1-\mathrm{\alpha }}{\mathrm{\alpha }}$ and $\frac{1-\mathrm{\beta }}{\mathrm{\beta }}$ then r=

$\begin{array}{l}Let\alpha ,\beta arerootsoff\left(x\right)=a{x}^2+bx+c=0\\ ThentheEquationwhoserootsare\\ \frac{1-\alpha }{\alpha }and\frac{1-\beta }{\beta }isf\left(\frac{1}{1-x}\right)=0\\ \Rightarrow a{\left(\frac{1}{1-x}\right)}^2+b\left(\frac{1}{1-x}\right)+c=0\\ \Rightarrow c{(1-x)}^2+b(1-x)+a=0\\ \Rightarrow c(1+x^2-2x)+b-bx+a=0\\ \Rightarrow c{x}^2-(2c+b)x+(a+b+c)=0\\ comparingwithp{x}^2+qx+c=0\\ p=c,q=-(2c+b),r=a+b+c\end{array}$