If the ratio of the roots of x^2 + bx + c = 0 | vedclass

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(C) Let $\alpha, \beta$ be the roots of $x^2 + bx + c = 0$ and $\alpha', \beta'$ be the roots of $x^2 + qx + r = 0$.Then $\alpha + \beta = -b, \alpha\

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$rb^2 = cq^2$

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(C) Let $\alpha, \beta$ be the roots of $x^2 + bx + c = 0$ and $\alpha', \beta'$ be the roots of $x^2 + qx + r = 0$.Then $\alpha + \beta = -b, \alpha\beta = c$ and $\alpha' + \beta' = -q, \alpha'\beta' = r$.Given that $\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'}$,we apply the property of componendo and dividendo: $\frac{\alpha + \beta}{\alpha - \beta} = \frac{\alpha' + \beta'}{\alpha' - \beta'}$.Squaring both sides,we get $\frac{(\alpha + \beta)^2}{(\alpha - \beta)^2} = \frac{(\alpha' + \beta')^2}{(\alpha' - \beta')^2}$.Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$,we have $\frac{b^2}{b^2 - 4c} = \frac{q^2}{q^2 - 4r}$.Cross-multiplying gives $b^2(q^2 - 4r) = q^2(b^2 - 4c) \Rightarrow b^2q^2 - 4rb^2 = q^2b^2 - 4cq^2$.Simplifying,we get $-4rb^2 = -4cq^2$,which implies $rb^2 = cq^2$.

If the ratio of the roots of $x^2 + bx + c = 0$ and $x^2 + qx + r = 0$ is the same,then

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