If the roots of the equation $a x^{2} + b x + c = 0$ be in the ratio 3:4, show that $12 b^{2} = 49 a c$ .

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Solution

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$x = \frac{- b \pm \sqrt{D}}{2 a}$

$d^{2} = b^{2} - 4 a c \Rightarrow d = \pm \sqrt{D}$

$x = - \frac{b}{2 a} \pm \frac{d}{2 a}$

$x_{1} = - \frac{b}{2 a} + \frac{d}{2 a}$

$x_{2} = - \frac{b}{2 a} - \frac{d}{2 a}$

Given,

$\frac{x_{1}}{x_{2}} = \frac{3}{4}$

$\frac{d - b}{- (b + d)} = \frac{3}{4}$

$4 (d - b) = 3 [- (b + d)]$

$4 d - 4 b = - 3 b - 3 d$

$b = 7 d$

$\Rightarrow b^{2} = 49 d^{2}$

$b^{2} = 49 (b^{2} - 4 a c)$

$48 b^{2} = 196 a c$

$12 b^{2} = 49 a c$

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