If 2 and 6 are the roots of the equation ax^2 | vedclass

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(D) Given the roots of $ax^2 + bx + 1 = 0$ are $2$ and $6$. Sum of roots: $2 + 6 = 8 = -\frac{b}{a} \implies b = -8a$. Product of roots: $2 	imes 6 =

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$x^2 + 8x + 12 = 0$

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(D) Given the roots of $ax^2 + bx + 1 = 0$ are $2$ and $6$. Sum of roots: $2 + 6 = 8 = -\frac{b}{a} \implies b = -8a$. Product of roots: $2 	imes 6 = 12 = \frac{1}{a} \implies a = \frac{1}{12}$. Substituting $a$ into $b = -8a$: $b = -8 	imes \frac{1}{12} = -\frac{2}{3}$. Now,calculate the new roots: Root $1 = \frac{1}{2a + b} = \frac{1}{2(\frac{1}{12}) - \frac{2}{3}} = \frac{1}{\frac{1}{6} - \frac{4}{6}} = \frac{1}{-\frac{3}{6}} = -2$. Root $2 = \frac{1}{6a + b} = \frac{1}{6(\frac{1}{12}) - \frac{2}{3}} = \frac{1}{\frac{1}{2} - \frac{2}{3}} = \frac{1}{-\frac{1}{6}} = -6$. The quadratic equation with roots $-2$ and $-6$ is $(x + 2)(x + 6) = 0$. $x^2 + 8x + 12 = 0$.

If $2$ and $6$ are the roots of the equation $ax^2 + bx + 1 = 0$,then the quadratic equation,whose roots are $\frac{1}{2a + b}$ and $\frac{1}{6a + b}$,is:

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