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

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Solution

$a x^{2} + b x + c = 0$
roots are $\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}$
$α = \frac{- b + \sqrt{b^{2} 4 a c}}{2 a} β = \frac{- b - \sqrt{b^{2} - 4 a x}}{2 a}$
$r = \frac{α}{β} = (\frac{- b + \sqrt{b^{2} 4 a c}}{- b + \sqrt{b^{2} - 4 a c}})$
$= \frac{b - \sqrt{b^{2} - 4 a c}}{- (b + \sqrt{b^{2} - 4 a c})} \times \frac{b - \sqrt{b^{2} - 4 a c}}{b - \sqrt{b^{2} - 4 a c}}$
$= \frac{(b - \sqrt{b^{2} - 4 a c})^{2}}{b - b^{2} + 4 a c}$
$= \frac{b^{2} + b^{2} - 4 a c - 2 b \sqrt{b^{2} - 4 a c}}{4 a c}$
$= \frac{2 b^{2} - 2 b \sqrt{b^{2} - 4 a c} - 4 a c}{4 a c}$

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