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

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(A) Let the roots of the equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.From the properties of quadratic equations,we have $\alpha + \beta = -\f

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(A) Let the roots of the equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.From the properties of quadratic equations,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.The sum of the reciprocals of their squares is given by $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$.Since $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$,we substitute the values:$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2}$.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 both sides by $abc$,we get $\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$ (This is not the standard form,let's look at the sequence $bc^2, ca^2, ab^2$).For $bc^2, ca^2, ab^2$ to be in $A.P.$,the middle term must be the average of the other two: $2(ca^2) = bc^2 + ab^2$.This matches our derived equation $2a^2c = ab^2 + bc^2$.Thus,$bc^2, ca^2, ab^2$ are 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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