If alpha and beta are the roots of ax^2 + bx | vedclass

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(D) Given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Let the n

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$ax^2 + x(b - 4a) + 4a - 2b + c = 0$

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(D) Given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Let the new roots be $\alpha' = 2 + \alpha$ and $\beta' = 2 + \beta$.The sum of the new roots is $\alpha' + \beta' = (2 + \alpha) + (2 + \beta) = 4 + (\alpha + \beta) = 4 - \frac{b}{a} = \frac{4a - b}{a}$.The product of the new roots is $\alpha'\beta' = (2 + \alpha)(2 + \beta) = 4 + 2(\alpha + \beta) + \alpha\beta = 4 + 2(-\frac{b}{a}) + \frac{c}{a} = \frac{4a - 2b + c}{a}$.The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - (\frac{4a - b}{a})x + \frac{4a - 2b + c}{a} = 0$.Multiplying by $a$,we obtain $ax^2 - (4a - b)x + 4a - 2b + c = 0$,which simplifies to $ax^2 + x(b - 4a) + 4a - 2b + c = 0$.

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$,then the equation whose roots are $2 + \alpha$ and $2 + \beta$ is:

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