If a polynomial equation x^3 - 6x^2 + 11x - 6 = 0 has roots r_1, r_2, and r_3, what is the sum of the squares of the roots: r_1^2 + r_2^2 + r_3^2?

Answer and explanation

Correct answer: C. 14

  1. Step 1: By Vieta's formulas for a cubic equation x^3 - a_2 x^2 + a_1 x - a_0 = 0:
  2. Step 2: r_1 + r_2 + r_3 = 6
  3. Step 3: r_1 r_2 + r_2 r_3 + r_1 r_3 = 11
  4. Step 4: The algebraic identity for the sum of squares is: (r_1 + r_2 + r_3)^2 = r_1^2 + r_2^2 + r_3^2 + 2(r_1 r_2 + r_2 r_3 + r_1 r_3).
  5. Step 5: Substitute the known values: 6^2 = (r_1^2 + r_2^2 + r_3^2) + 2(11).
  6. Step 6: 36 = (r_1^2 + r_2^2 + r_3^2) + 22.
  7. Step 7: r_1^2 + r_2^2 + r_3^2 = 36 - 22 = 14.
Answer
14.
Why not the others: '36' is the square of the sum (6^2); '22' is twice the pairwise product sum; '12' is 36 - 2(12).

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