Question

If $r$ be the ratio of the roots of the equation $a{x}^2+bx+c=0$, show that $r{b}^2={(r+1)}^{2}ac$

Answer 1

$a{x}^2+bx+c=0\to$ roots, $\alpha ,\beta$

$\frac{\alpha }{\beta }=x$

using componundo - divoderdo

$\frac{\alpha +\beta }{\alpha -\beta }=\frac{r+1}{r-1}$

$\frac{-b/a}{\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }}=\frac{r+1}{r-1}$

$\frac{-b/a}{\sqrt{b^2/a^2-4c/a}}=\frac{r+1}{r-1}\Rightarrow \frac{-b/a}{\sqrt{\frac{b^2-4ac}{a}}}=\frac{r+1}{r-1}$

$b^2(r-1)=(r+1{)}^2(b^2-4ac)$

$b^2(x-1{)}^2=(r+1{)}^2{b}^2-(r+1{)}^{2}4ac$

$(r+1{)}^{2}4ac=((r+1{)}^2-(x-1{)}^2)b^2=4r{b}^2$

$(r+1{)}^{2}ac=r{b}^2$