If alpha and beta are the roots of the equati | vedclass

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(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$a

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$-2/a$

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(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$a\alpha^2 + b\alpha + c = 0 \implies a\alpha^2 + b\alpha = -c \implies \alpha(a\alpha + b) = -c \implies a\alpha + b = -c/\alpha$.Similarly,$a\beta^2 + b\beta + c = 0 \implies a\beta^2 + b\beta = -c \implies \beta(a\beta + b) = -c \implies a\beta + b = -c/\beta$.Now,substitute these into the expression:$\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b} = \frac{\alpha}{-c/\beta} + \frac{\beta}{-c/\alpha}$.$= -\frac{\alpha\beta}{c} - \frac{\alpha\beta}{c} = -\frac{2\alpha\beta}{c}$.Since the product of roots $\alpha\beta = c/a$,we substitute this value:$= -\frac{2}{c} 	imes \frac{c}{a} = -\frac{2}{a}$.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then $\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b} = \dots$

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