Multiple choice

During a math competition, Garry and Daniel were given a polynomial problem to solve. If one zero of the polynomial f(x) = x2 - (k - 3)x + 12 is triple the other zero, what is the sum of all the possible values of k?

  1. -11

  2. 6

  3. 7

  4. 9

  5. 11
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B Correct answer
Explanation

Let roots be a and 3a. Sum of roots = 4a = k - 3. Product of roots = 3a^2 = 12, so a^2 = 4, a = 2 or -2. If a = 2, 4(2) = k - 3 => k = 11. If a = -2, 4(-2) = k - 3 => k = -5. Sum of possible k values = 11 + (-5) = 6.

AI explanation

Let the zeros be a and 3a; using Vieta's formulas for f(x) = x^2 - (k - 3)x + 12, the sum of zeros gives a + 3a = k - 3 and the product of zeros gives a * 3a = 12. Solving the product equation yields 3a^2 = 12, so a = 2 or a = -2. If a = 2, the sum is 8, making k = 11; if a = -2, the sum is -8, making k = -5. Adding these possible values of k gives 11 + (-5) = 6.

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