For which range of values of k does the quadratic equation x^2 - 2(k - 1)x + (k + 5) = 0 have no real roots?

Answer and explanation

Correct answer: C. -1 < k < 4

  1. Step 1: An equation has no real roots when its discriminant Delta < 0.
  2. Step 2: Delta = B^2 - 4AC = [-2(k - 1)]^2 - 4(1)*(k + 5).
  3. Step 3: Delta = 4(k^2 - 2k + 1) - 4(k + 5) = 4(k^2 - 3*k - 4).
  4. Step 4: Set 4(k^2 - 3k - 4) < 0, so k^2 - 3*k - 4 < 0.
  5. Step 5: Factor: (k - 4)(k + 1) < 0.
  6. Step 6: This quadratic is strictly negative strictly between its roots: -1 < k < 4.
Answer
-1 < k < 4.
Why not the others: k < -1 or k > 4 gives Delta > 0 (two distinct real roots); -4 < k < 1 has reversed root signs; k > 4 only is one-sided.

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