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(B) Given that $\alpha, \beta$ are the roots of $Ax^2 + Bx + C = 0$.By the relation between roots and coefficients,$\alpha + \beta = -\frac{B}{A}$ and

Vedclass · Accepted Answer

$\frac{2AC - B^2}{A^2}$

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(B) Given that $\alpha, \beta$ are the roots of $Ax^2 + Bx + C = 0$.By the relation between roots and coefficients,$\alpha + \beta = -\frac{B}{A}$ and $\alpha\beta = \frac{C}{A}$.Given that $\alpha^2, \beta^2$ are the roots of $x^2 + px + q = 0$.By the relation between roots and coefficients,$\alpha^2 + \beta^2 = -p$ and $\alpha^2\beta^2 = q$.We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.Substituting the values,we get $\alpha^2 + \beta^2 = (-\frac{B}{A})^2 - 2(\frac{C}{A}) = \frac{B^2}{A^2} - \frac{2C}{A} = \frac{B^2 - 2AC}{A^2}$.Since $\alpha^2 + \beta^2 = -p$,we have $-p = \frac{B^2 - 2AC}{A^2}$,which implies $p = \frac{2AC - B^2}{A^2}$.

If the roots of the equation $Ax^2 + Bx + C = 0$ are $\alpha, \beta$ and the roots of the equation $x^2 + px + q = 0$ are $\alpha^2, \beta^2$,then the value of $p$ will be

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