Multiple choice

The number of integers k for which the equation x3 - 27x + k = 0 has at least two distinct integer roots is

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

x^3 - 27x + k = 0. For at least two distinct integer roots, the function must have integer values for x. Testing values, k must be such that the cubic factors into (x-a)(x-b)(x-c). For integer roots, the product of roots is -k and sum is 0. Possible integer roots for x^3-27x+k=0: if roots are a, b, c, then a+b+c=0 and ab+bc+ca=-27. If a=3, b=3, c=-6, k = -(3*3*-6) = 54. If a=-3, b=-3, c=6, k = -(-3*-3*6) = -54. Two values.

AI explanation

Let the integer roots of the cubic equation x^3 - 27x + k = 0 be r, s, and t; any integer root must divide the constant term k. By Vieta's formulas, the sum of the roots is zero (r + s + t = 0) and the sum of their pairwise products is -27 (rs + rt + st = -27). Testing small integer combinations that sum to zero reveals the roots can be (-6, 3, 3), which gives k = 54, and (-3, -3, 6), which gives k = -54. Therefore, there are exactly 2 possible integer values for k.

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