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(B) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $l$ and $2l$.Sum of roots: $l + 2l = -\frac{b}{a}$ $\Rightarrow 3l = -\frac{b}{a}$ $\R

Vedclass · Accepted Answer

$2b^2 = 9ac$

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(B) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $l$ and $2l$.Sum of roots: $l + 2l = -\frac{b}{a}$ $\Rightarrow 3l = -\frac{b}{a}$ $\Rightarrow l = -\frac{b}{3a}$ $(i)$Product of roots: $l 	imes 2l = \frac{c}{a} \Rightarrow 2l^2 = \frac{c}{a}$ $(ii)$Substituting $(i)$ into $(ii)$:$2\left(-\frac{b}{3a}ight)^2 = \frac{c}{a}$$2\left(\frac{b^2}{9a^2}ight) = \frac{c}{a}$$\frac{2b^2}{9a^2} = \frac{c}{a}$$2b^2 = 9ac$Alternatively,for a quadratic equation $ax^2 + bx + c = 0$ with roots in ratio $m:n$,the condition is $mnb^2 = (m+n)^2ac$. Here $m=1, n=2$,so $1 	imes 2 	imes b^2 = (1+2)^2ac \Rightarrow 2b^2 = 9ac$.

If the roots of the equation $ax^2 + bx + c = 0$ are $l$ and $2l$,then

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