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14 October, 05:04

For what values of k does the equation x^2 - (5k+1) x + 9k^2 = 0 have real roots?

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Answers (2)
  1. 14 October, 05:37
    0
    K has a range from[ - 1/11 to 1]
  2. 14 October, 06:51
    0
    Real roots occur when the determinant b^2-4ac > = 0

    so substituting:-

    (5k + 1) ^2 - 4*1 * (9k^2) > = 0

    25k^2 + 10k + 1 - 36k^2 > = 0

    -11k^2 + 10k + 1 > = 0

    11k^2 - 10k - 1 < = 0

    (11k + 1) (k - 1) < = 0

    the equation will have real roots for - 1/11 = < k < = 1
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