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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$rb^2 = cq^2$

Vedclass · Answer

(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,$\alpha + \beta = -b$,$\alpha\beta = c$,$\alpha' + \beta' = -q$,and $\alpha'\beta' = r$.Given that the ratio of the roots is the same,$\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'}$.Using the property of componendo and dividendo,$\frac{\alpha + \beta}{\alpha - \beta} = \frac{\alpha' + \beta'}{\alpha' - \beta'}$.Squaring both sides,$\frac{(\alpha + \beta)^2}{(\alpha - \beta)^2} = \frac{(\alpha' + \beta')^2}{(\alpha' - \beta')^2}$.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 $b^2(q^2 - 4r) = q^2(b^2 - 4c)$.$b^2q^2 - 4b^2r = q^2b^2 - 4cq^2$.$-4b^2r = -4cq^2$,which simplifies to $b^2r = cq^2$.

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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