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21 July, 19:31

If $x^2+bx+9$ has two non-real roots, find all possible values of $b$. express your answer in interval notation.

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Answers (2)
  1. 21 July, 20:38
    0
    Applying the resolver we have:

    x = ( - b + / - Root (b ^ 2 - 4 * a * c)) / (2 * a)

    Substituting values we have:

    x = ( - b + / - Root (b ^ 2 - 4 * 1 * 9)) / (2 * 1)

    Rewriting:

    x = ( - b + / - Root (b ^ 2 - 36)) / (2)

    For non-real values we have:

    b ^ 2 - 36 <0

    We cleared b:

    b ^ 2 <36

    b < + / - root (36)

    b < + / - 6

    Rewriting:

    -6
    In interval notation:

    (-6, 6)

    Answer:

    all possible values of b are:

    (-6, 6)
  2. 21 July, 21:29
    0
    In order to have non-real roots, the radicand must be less than zero, therefore:

    x² + bx + 9 = 0

    will have as a radicand b² - 4·1·9

    Therefore:

    b² - 36 < 0

    First, find the solutions to the equation b² - 36 = 0

    which are b = + / - 6

    Since the disequality asks for "less than", we need to take the inner interval between these two solutions:

    -6 < b < + 6

    or else, ] - 6; + 6 [

    or else, (-6; + 6)

    (These are all valid notations).
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