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

<p>Given that the roots of the quadratic equation are $2$ and $6$, we can use Vieta's formulas which relate the coefficients of the polynomial to sums and products of its roots.</p> <p>The given quadratic equation is:</p> <p>$a x^2 + b x + 1 = 0$</p> <p>By Vieta's formulas, the sum of the roots is:</p> <p>$2 + 6 = -\frac{b}{a}$</p> <p>So:</p> <p>$8 = -\frac{b}{a} \Rightarrow b = -8a$</p> <p>And the product of the roots is:</p> <p>$$2 \times 6 = \frac{1}{a} \Rightarrow 12 = \frac{1}{a} \Rightarrow a = \frac{1}{12}$$</p> <p>Therefore, $b = -8a = -8 \left( \frac{1}{12} \right) = -\frac{2}{3}$.</p> <p>Given the roots of the new quadratic equation are:</p> <p>$\frac{1}{2a+b}$ and $\frac{1}{6a+b}$</p> <p>We know $a = \frac{1}{12}$ and $b = -\frac{2}{3}$, so:</p> <p>$$2a + b = 2 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = \frac{1 - 4}{6} = -\frac{3}{6} = -\frac{1}{2}$$</p> <p>and:</p> <p>$$6a + b = 6 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = \frac{3 - 4}{6} = -\frac{1}{6}$$</p> <p>Thus, the roots of the new quadratic equation are:</p> <p>$\frac{1}{-\frac{1}{2}} = -2$</p> <p>and:</p> <p>$\frac{1}{-\frac{1}{6}} = -6$</p> <p>The new quadratic equation with roots $-2$ and $-6$ can be formulated as:</p> <p>$x^2 - (\text{sum of roots}) x + (\text{product of roots}) = 0$</p> <p>The sum of the roots is:</p> <p>$-2 + (-6) = -8$</p> <p>The product of the roots is:</p> <p>$(-2) \times (-6) = 12$</p> <p>Thus, the quadratic equation becomes:</p> <p>$x^2 - (-8)x + 12 = x^2 + 8x + 12 = 0$</p> <p>Hence, the correct option is:</p> <p>Option A</p> <p>$x^2 + 8 x + 12 = 0$</p>