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14 February, 10:12

What is true about the solutions of a quadratic equation when the radicand of the quadratic formula is a perfect square?

Answer

No real solutions

Two identical rational solutions

Two different rational solutions

Two irrational solutions

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Answers (1)
  1. 14 February, 12:39
    0
    The solution of quadratic formula is represented as

    x = - b ± squareroot (b^2 - 4ac) / 2a

    Where b^2 - 4ac is known as the radicand.

    Lets take an equation; x^2 + 6x + 5 = 0

    Where a = 1, b = 6, c = 5

    The solution will be x = - 6 ± squareroot (6^2 - 4 (1) (5)) / 2 (1)

    Where radicand = 6^2 - 4 (1) (5) = 36 - 20 = 16

    Hence radicand is the perfect square of 4. i. e. 4^2 = 16

    Now solving the equation further;

    x = (-6 ± 4) / 2

    x = - 1, x = - 5

    Thus for a radicand to be a perfect square, the solution of a quadratic equation will composed of two different rational solutions.
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