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x(6x+1)<15
can some one explain the problem and answer to me step by step please.. I missing the process

  • math -

    x(6x+1) < 15
    6x^2 + x - 15 < 0
    now you have a parabola. You need to find where it crosses the x-axis, and between those two points, it will lie below the x-axis.

    (2x-3)(3x+5) < 0

    so, the roots are at x = -5/3 and 3/2

    Between those two values, the parabola lies below the x-axis.

    As a check, let's try the original conditions.

    x(6x+1) < 15
    at x = -5/3, x(6x+1) = 15
    at x = 3/2, x(6x+1) = 15
    Now check something in between, say x = -1,0,1
    f(x) = 5,0,7 all < 15.

    How about a few outside -5/3 < x < 3/2 ?
    f(-2) = 22 > 15
    f(2) = 26 > 15

    Take a look here:

    http://www.wolframalpha.com/input/?i=x%286x%2B1%29+%3C+15

  • math -

    Initial Problem
    x(6x + 1) < 15

    Simplify left side.
    6x^2 + x < 15

    Move 15 to left side.
    6x^2 + x - 15 < 0

    Factor.
    (2x - 3)(3x + 5) < 0

    Now, x MUST be BETWEEN the LESSER value and the GREATER value. (it is between because this is a "LESS THAN" inequality)
    -5/3 < x < 3/2

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