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8 December, 06:32

If c is a real number and if 1 + i is a solution of the equation x^2 - 2x + c = 0, what is the value of c?

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Answers (2)
  1. 8 December, 06:50
    0
    c = 2.

    Step-by-step explanation:

    Complex solutions come in conjugate pairs so the other solution is 1 - i.

    So we have:

    (x - (1 + i) (x - (1 - i) = 0

    (x - 1 - i) (x - 1 + i) = 0

    x^2 - x + ix - x + 1 - i - ix + i - i^2 = 0

    x^2 - 2x + 1 - i^2 = 0

    x^2 - 2x + 1 + 1 = 0

    x^2 - 2x + 2 = 0

    c = 2.
  2. 8 December, 06:57
    0
    c = 2

    Step-by-step explanation:

    Substitute x = 1 + i into the equation

    (1 + i) ² - 2 (1 + i) + c = 0 ← distribute left side

    1 + 2i + i² - 2 - 2i + c = 0 (note that i² = - 1)

    1 + 2i - 1 - 2 - 2i + c = 0 ← collect like terms on left

    - 2 + c = 0 (add 2 to both sides)

    c = 2
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