Question

Let $\alpha$ and $\beta$ are the roots of the equation $a{x}^2+bx+c=0.$ If $\frac{1-\alpha }{\alpha }$ and $\frac{1-\beta }{\beta }$ are the roots of $p{x}^2+qx+r=0$ and $\frac{q}{p}=k+\frac{b}{c}$, then the value of $k$ is

Answer 1

Let $f\left(x\right)=a{x}^2+bx+c=0$ has roots $\alpha ,\beta$
$y=\frac{1-x}{x}$, where $x=\alpha ,\beta$
$y=\frac{1}{x}-1\Rightarrow y+1=\frac{1}{x}\Rightarrow x=\frac{1}{y+1}$

The equation whose $\frac{1-\alpha }{\alpha }$ and $\frac{1-\beta }{\beta }$ is,
$f\left(\frac{1}{x+1}\right)=0\Rightarrow a+b(x+1)+c(x+1{)}^2=0\Rightarrow c{x}^2+(2c+b)x+(a+b+c)=0$
Comparing with
$p{x}^2+qx+r=0$
$\frac{c}{p}=\frac{(2c+b)}{q}=\frac{(a+b+c)}{r}=k$
Now,
$\frac{c}{p}=\frac{(2c+b)}{q}$
$\Rightarrow \frac{q}{p}=\frac{(2c+b)}{c}\Rightarrow \frac{q}{p}=2+\frac{b}{c}\therefore k=2$