If the sum of the roots of the equation ax^2 | vedclass

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(A) Given the equation $ax^2 + bx + c = 0$. Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}

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$A.P.$

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(A) Given the equation $ax^2 + bx + c = 0$. Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.The sum of the reciprocals of their squares is $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.Substituting the values: $\frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2}$.According to the given condition,$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$,so $-\frac{b}{a} = \frac{b^2 - 2ac}{c^2}$.Cross-multiplying gives $-bc^2 = a(b^2 - 2ac) = ab^2 - 2a^2c$.Rearranging the terms,we get $2a^2c = ab^2 + bc^2$.Dividing by $a^2c^2$ is not necessary here; we observe that $2(ca^2) = bc^2 + ab^2$. This is the condition for $bc^2, ca^2, ab^2$ to be in $A.P.$

If the sum of the roots of the equation $ax^2 + bx + c = 0$ is equal to the sum of the reciprocals of their squares,then $bc^2, ca^2, ab^2$ will be in

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