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3 October, 21:36

B. x4 + 10x2 + 25 = 0

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  1. 3 October, 21:48
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    Solve it like a normal quadratic equation and take the root of the roots

    x^4 + 10x^2 + 25 = 0

    Δ = b² - 4. a. c

    Δ = 10² - 4. 1. 25

    Δ = 100 - 4. 1. 25

    Δ = 0

    1 real root.

    In this case, x' = x'':

    x = (-b + - √Δ) / 2a

    x' = (-10 + √0) / 2.1

    x'' = (-10 - √0) / 2.1

    x' = - 10 / 2

    x'' = - 10 / 2

    x' = - 5

    x'' = - 5

    Now take the root of this roots

    √-5 = i√5

    The root of this equation is x = ±i√5
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