Let f(x)=a x^2+b x+c and the GCD of a, b, c b | vedclass

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(B) Given $f(x)=a x^2+b x+c$ and $GCD(a, b, c)=1$. Since $\frac{-7+\sqrt{11} i}{6}$ is a root of $f(x)=0$,its conjugate $\frac{-7-\sqrt{11} i}{6}$ mus

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$1, 25$

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(B) Given $f(x)=a x^2+b x+c$ and $GCD(a, b, c)=1$. Since $\frac{-7+\sqrt{11} i}{6}$ is a root of $f(x)=0$,its conjugate $\frac{-7-\sqrt{11} i}{6}$ must also be a root. Sum of roots $= \frac{-7+\sqrt{11} i}{6} + \frac{-7-\sqrt{11} i}{6} = \frac{-14}{6} = -\frac{7}{3} = -\frac{b}{a} \implies \frac{b}{a} = \frac{7}{3}$. Product of roots $= \left(\frac{-7+\sqrt{11} i}{6}\right) \left(\frac{-7-\sqrt{11} i}{6}\right) = \frac{(-7)^2 - (\sqrt{11} i)^2}{36} = \frac{49+11}{36} = \frac{60}{36} = \frac{5}{3} = \frac{c}{a}$. Thus,$a=3, b=7, c=5$,which satisfies $GCD(3, 7, 5)=1$. So,$f(x) = 3x^2+7x+5$. Given $f\left(\frac{x}{k}\right)-L = (x+4)(3x-5) = 3x^2+7x-20$. Substituting $f\left(\frac{x}{k}\right) = 3\left(\frac{x}{k}\right)^2 + 7\left(\frac{x}{k}\right) + 5$,we get: $3\frac{x^2}{k^2} + \frac{7x}{k} + 5 - L = 3x^2 + 7x - 20$. Comparing coefficients: $\frac{3}{k^2} = 3 \implies k^2 = 1 \implies k = 1$ (taking positive root). $\frac{7}{k} = 7 \implies k = 1$. $5 - L = -20 \implies L = 25$. Therefore,$k=1$ and $L=25$.

Let $f(x)=a x^2+b x+c$ and the $GCD$ of $a, b, c$ be $1$. If $\frac{-7+\sqrt{11} i}{6}$ is a root of $f(x)=0$ and $f\left(\frac{x}{k}\right)-L=(x+4)(3 x-5)$,then $k$ and $L$ are respectively:

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