Multiple choice

The equation of smallest degree with real coefficients having 2 + 3i as one of the roots is

  1. x2 − 4x + 13 = 0

  2. x2 + 5x + 6 = 0

  3. x2 − 2x + 1 = 0

  4. x2 + 2x + 1 = 0
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A Correct answer
Explanation

If 2+3i is a root, 2-3i must also be a root. The equation is (x - (2+3i))(x - (2-3i)) = ((x-2)-3i)((x-2)+3i) = (x-2)^2 + 9 = x^2 - 4x + 4 + 9 = x^2 - 4x + 13 = 0.

AI explanation

For a polynomial with real coefficients, any complex root must be accompanied by its conjugate, making the second root 2 - 3i. The required polynomial is found by multiplying the linear factors (x - (2 + 3i)) and (x - (2 - 3i)) to get x2 - 4x + 13. This gives the equation x2 - 4x + 13 = 0.

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