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Solving inequalities, need some confirmation

  1. Jul 27, 2005 #1
    I have these three inequalities that I am supposed to solve, I think I came up with the right answer but I'm not even 100% sure it's in the correct format.

    A. 6x^2 < 6+5x
    my work:
    6x^2-5x-6 < 0
    solutions are then 3/2 and -2/3
    so the answer I got is:
    -2/3 < x < 3/2

    B. x^2+8x > 0
    my work:
    soutions I got were 0, -8
    so my answer is:
    -8 < x < 0

    C. (x+2)(x^2-x+1) > 0
    my work:
    x+2 > 0 and x^2-x+1 > 0
    solutions are then -.414 and 2.414
    so my answer is:
    -4.14 < x < 2.414
  2. jcsd
  3. Jul 27, 2005 #2
    Your first one looks fine....

    [tex]y = x^2+8x[/tex]

    Opens up...the vertex is at (-4, -16)...you found the zeros at 0 and -8.
    Graph this....Now find the parts of the graph that are BIGGER than 0...in other words, what parts of the graph are above the x-axis?

    This has only one zero....x = -2.
    x^2 - x +1 = (x - 1/2)^2 + 3/4

    Which is clearly always above the x-axis and so you have no real roots....

    So to the left of ( -2, 0)...for example x = -3, what would your expression evaluate to? A positive or negative number?
  4. Jul 27, 2005 #3


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    No, no no!
    I'll take B for you:
    This can be rewritten as:

    What you call "solutions", are the solutions to the equation [tex]x^{2}+8x=0[/tex]
    These values are important in determining the regions of x-values where the INEQUALITY holds, but they are by no means indicative of these regions in the manner you think.

    Let us go back to:
    The left-hand side has two factors.
    The product of two numbers are positive if
    a) each factor is positive (that is, x>0 AND, x+8>0)
    b) each factor is negative (that is, x<0 AND x+8<0)

    Try now to identify the regions on the x-axis where the inequality holds!
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