The sum of the roots of a quadratic equation | vedclass

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(D) Let the roots of the quadratic equation be $\alpha$ and $\beta$.Given $\alpha + \beta = 2$ and $\alpha^3 + \beta^3 = 98$.We know that $\alpha^3 +

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$x^2 - 2x - 15 = 0$

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(D) Let the roots of the quadratic equation be $\alpha$ and $\beta$.Given $\alpha + \beta = 2$ and $\alpha^3 + \beta^3 = 98$.We know that $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$.This can be written as $\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta)$.Substituting the given values: $98 = 2((2)^2 - 3\alpha\beta)$.$49 = 4 - 3\alpha\beta$.$3\alpha\beta = 4 - 49 = -45$.$\alpha\beta = -15$.The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Therefore,$x^2 - 2x - 15 = 0$.

The sum of the roots of a quadratic equation is $2$ and the sum of their cubes is $98$. Then the equation is:

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