We have,
$\alpha, \beta$ are the roots of the equations $a x^2+b x+c=0$
$$
\therefore \quad \alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}
$$
Equations whose roots are $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$ are
$$
\begin{aligned}
& x^2-\left(\frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}\right) x+\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=0 \\
& \Rightarrow \quad x^2-\left(\frac{1}{\alpha}+\frac{1}{\beta}-2\right) x+\left(\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\right)=0 \\
& \Rightarrow x^2-\left(\frac{\alpha+\beta-2 \alpha \beta}{\alpha \beta}\right) x+\left(\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\right)=0 \\
& \Rightarrow \alpha \beta x^2-(\alpha+\beta-2 \alpha \beta) x+(1-\langle\alpha+\beta\rangle+\alpha \beta)=0 \\
& \Rightarrow \quad \frac{c}{a} x^2-\left(-\frac{b}{a}-\frac{2 c}{a}\right) x+\left(1+\frac{b}{a}+\frac{c}{a}\right)=0 \\
& \Rightarrow \quad a x^2+(b+2 c) x+(a+b+c)=0 \\
&
\end{aligned}
$$
Comparing with $p x^2+q x+r=0$, we get
$$
r=a+b+c
$$