If the sum of the roots of a quadratic equati | vedclass

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(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$. Given that the sum of the roots is $\alpha + \beta = -1$. Given that the sum of t

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$x^2 + x - 6 = 0$

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(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$. Given that the sum of the roots is $\alpha + \beta = -1$. Given that the sum of their reciprocals is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{6}$. This simplifies to $\frac{\alpha + \beta}{\alpha \beta} = \frac{1}{6}$. Substituting $\alpha + \beta = -1$,we get $\frac{-1}{\alpha \beta} = \frac{1}{6}$,which implies $\alpha \beta = -6$. The quadratic equation is given by $x^2 - (\alpha + \beta)x + \alpha \beta = 0$. Substituting the values,we get $x^2 - (-1)x + (-6) = 0$,which simplifies to $x^2 + x - 6 = 0$.

If the sum of the roots of a quadratic equation is $-1$ and the sum of their reciprocals is $\frac{1}{6}$,then the equation is:

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