Sum of roots is -1 and sum of their reciproca | 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 the

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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 $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{6}$.Simplifying the sum of reciprocals: $\frac{\alpha + \beta}{\alpha \beta} = \frac{1}{6}$.Substituting $\alpha + \beta = -1$ into the equation: $\frac{-1}{\alpha \beta} = \frac{1}{6}$,which gives $\alpha \beta = -6$.The standard form of a quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values: $x^2 - (-1)x + (-6) = 0$.Therefore,the equation is $x^2 + x - 6 = 0$.

Sum of roots is $-1$ and sum of their reciprocals is $\frac{1}{6}$,then the equation is

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