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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given that $\alpha, \beta$ are the roots of $Ax^2 + Bx + C = 0$. From the relation between roots and coefficients,$\alpha + \beta = -B/A$ and $\alpha\beta = C/A$. Given that $\alpha^2, \beta^2$ are the roots of $x^2 + px + q = 0$. From 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,$-p = (-B/A)^2 - 2(C/A)$. $-p = \frac{B^2}{A^2} - \frac{2C}{A} = \frac{B^2 - 2AC}{A^2}$. Therefore,$p = -\frac{B^2 - 2AC}{A^2} = \frac{2AC - B^2}{A^2}$.

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

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