If the ratio of the roots of ax^2 + 2bx + c = | vedclass

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Question

(B) Let the roots of $ax^2 + 2bx + c = 0$ be $\alpha$ and $\beta$ such that $\frac{\alpha}{\beta} = \frac{m}{n}$.Then $\alpha + \beta = -\frac{2b}{a}$

Vedclass · Accepted Answer

$\frac{b^2}{ac} = \frac{q^2}{pr}$

Vedclass · Answer

(B) Let the roots of $ax^2 + 2bx + c = 0$ be $\alpha$ and $\beta$ such that $\frac{\alpha}{\beta} = \frac{m}{n}$.Then $\alpha + \beta = -\frac{2b}{a}$ and $\alpha\beta = \frac{c}{a}$.Using the relation $\frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{(\frac{m}{n} + 1)^2}{\frac{m}{n}} = \frac{(m+n)^2}{mn}$,we get $\frac{4b^2/a^2}{c/a} = \frac{(m+n)^2}{mn}$,which simplifies to $\frac{4b^2}{ac} = \frac{(m+n)^2}{mn}$.Similarly,for $px^2 + 2qx + r = 0$,we have $\frac{4q^2}{pr} = \frac{(m+n)^2}{mn}$.Equating the two,we get $\frac{4b^2}{ac} = \frac{4q^2}{pr}$,which implies $\frac{b^2}{ac} = \frac{q^2}{pr}$.

If the ratio of the roots of $ax^2 + 2bx + c = 0$ is the same as the ratio of the roots of $px^2 + 2qx + r = 0$,then

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