Please help with an inequality sum?

3/x + x/2 < 0

Solving it by algebraic method leaves me with no real solution. But when I plotted a graph using the GC, my answer is x <0.

Can anyone explain?

3 Answers

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  • 8 years ago
    Favorite Answer

    3/x + x/2 < 0

    [2(3) / 2x] + (x² / 2x) < 0

    (6 / 2x) + (x² / 2x) < 0

    (6 + x²) / 2x < 0

    If the quotient of two terms is negative, then one, and only one, term is negative.

    If 6 + x² < 0,

    x² < - 6

    x < √(- 6)

    x < √6 √(- 1)

    x < √6 i

    If 2x < 0,

    x < 0

    Since x < √6 i is imaginary, it does not exist, so

    x < 0

    ¯¯¯¯

     

    Source(s): 4/10/13
  • Josh K
    Lv 6
    8 years ago

    Multiply through by x

    What you need to realize is that x can be negative in some circumstances affecting the inequality.

    Case 1, x>0

    3 + 1/2 x^2 < 0

    x^2 < -6 Which is impossible in the real numbers, so this case yields no solution

    Case 2, x = 0 undefined in original equation as a denominator = 0

    Case 3 x < 0

    3 + 1/2 x^2 > 0

    1/2 x^2 > -3

    x^2 > -6

    Since x^2 will always be positive for any value of x, this will always be true.

    Then, to come up with the solution, the criteria for this case is x<0 AND x any real number.

    This intersection is identical to x < 0

    So here, the defining characteristic of the case actually becomes the solution.

  • 8 years ago

    Anybody can see that all negative values of x is solutions for the inequality.

    If we multiply both sides of the inequality by 2x we get (6+x^2)/2x < 0

    for any negative number (6+x^2) is positive and 2x is negative, and this inequality is also fullfilled for all negative values of x.

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