The number of integers k for which the equati | vedclass

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(B) Let $f(x) = x^3 - 27x + k = 0$. Let the roots be $\alpha, \beta, \gamma$. Since the coefficient of $x^2$ is $0$,the sum of the roots $\alpha + \be

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$2$

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(B) Let $f(x) = x^3 - 27x + k = 0$. Let the roots be $\alpha, \beta, \gamma$. Since the coefficient of $x^2$ is $0$,the sum of the roots $\alpha + \beta + \gamma = 0$. If two roots are distinct integers,the third root must also be an integer.Let the roots be $x_1, x_2, x_3$. Then $x_1 + x_2 + x_3 = 0$ and $x_1 x_2 + x_2 x_3 + x_3 x_1 = -27$. Substituting $x_3 = -(x_1 + x_2)$ into the second equation: $x_1 x_2 - (x_1 + x_2)^2 = -27$,which simplifies to $x_1^2 + x_1 x_2 + x_2^2 = 27$.This is the equation of an ellipse. We look for integer solutions $(x_1, x_2)$.If $x_1 = 0$,$x_2^2 = 27$ (no integer solution).If $x_1 = 3$,$9 + 3x_2 + x_2^2 = 27$ $\Rightarrow x_2^2 + 3x_2 - 18 = 0$ $\Rightarrow (x_2+6)(x_2-3) = 0$. So $x_2 = 3$ or $x_2 = -6$.If $x_2 = 3$,roots are $(3, 3, -6)$. Then $k = -(x_1 x_2 x_3) = -(3 	imes 3 	imes -6) = 54$.If $x_2 = -6$,roots are $(3, -6, 3)$,same as above.If $x_1 = -3$,$9 - 3x_2 + x_2^2 = 27$ $\Rightarrow x_2^2 - 3x_2 - 18 = 0$ $\Rightarrow (x_2-6)(x_2+3) = 0$. So $x_2 = 6$ or $x_2 = -3$.If $x_2 = 6$,roots are $(-3, 6, -3)$. Then $k = -(-3 	imes 6 	imes -3) = -54$.If $x_2 = -3$,roots are $(-3, -3, 6)$,same as above.Thus,$k = 54$ or $k = -54$. There are $2$ such integers.

$A. \alpha + \beta + \gamma$	$(1) -\frac{43}{4}$
$B. \alpha^2 + \beta^2 + \gamma^2$	$(2) -\frac{7}{8}$
$C. \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$	$(3) 86$
$D. \frac{\alpha}{\beta \gamma} + \frac{\beta}{\gamma \alpha} + \frac{\gamma}{\alpha \beta}$	$(4) 0$
	$(5) 10$

The number of integers $k$ for which the equation $x^3-27x+k=0$ has at least two distinct integer roots is

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