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Solve for x- the inequality of quadratic

  1. Aug 26, 2011 #1
    Solve for x-- the inequality of quadratic

    1. The problem statement, all variables and given/known data
    Solve [tex]\frac{2x}{x^2-9}\le\frac{1}{x+2}[/tex]



    3. The attempt at a solution
    [tex]x^2-9\not=0[/tex]
    .'. [tex]x\in R-\{-3,3\}[/tex]
    and

    [tex]x+2\not=0[/tex]
    .'. [tex]x\in R-\{-2\}[/tex]

    then converting the original inequality to
    [tex](2x)(x+2)\le(x^2-9)[/tex]
    [tex](2x^2+4x)(-x^2+9)\le 0[/tex]
    [tex](x^2+4x+9)\le 0[/tex]
    As Discriminant <0 it has no real roots
    so how to do further...




    My assumptions:-
    [tex]x^2-9[/tex]should not be zero
    [tex]x+2[/tex]should also not be zero

    Are my assumptions right?
    Any help will be highly appreciated.
     
  2. jcsd
  3. Aug 26, 2011 #2

    micromass

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    Re: Solve for x-- the inequality of quadratic

    I see what you did here, but it's not correct. It's not true that

    [tex]\frac{a}{b}\leq \frac{c}{d}[/tex]

    implies that [itex]ad\leq bc[/itex]. For example

    [tex]\frac{1}{-1}\leq \frac{1}{1}[/tex]

    but it's not true that [itex]1\leq -1[/itex]. The problem is of course that negative numbers reverse the inequality sign.

    How do we do such a thing then?? Well, first we move everything to the left side of the equation:

    [tex]\frac{2x}{x^2-9}-\frac{1}{x+2}\leq 0[/tex]

    and now you must try to add these fractions.
     
  4. Aug 26, 2011 #3

    Ray Vickson

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    Re: Solve for x-- the inequality of quadratic

    In general, if a <= b we have a*c <= b*c if c > 0 and a*c >= b*c if c < 0. Thus, if x+1 > 0 you have 2x(x+2)/(x^2-9) <= 1 and if x+2 < 0 you have the opposite inequality. Now, in each case, look at whether x^2-9 is positive or negative.

    RGV
     
  5. Aug 26, 2011 #4

    uart

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    Re: Solve for x-- the inequality of quadratic

    [tex]\frac{2x}{x^2-9}\le\frac{1}{x+2}[/tex]

    I often prefer to multiply by a square to avoid splitting into too many +/- cases. It's a compromise though, between getting a higher order system versus less case splitting.

    [tex]\frac{2x (x+2)^2}{x^2-9}\le x+2[/tex]

    I'll split into cases for the (x^2-9) factor to keep the order of the system under control.

    [tex]2x (x+2)^2 \le (x+2) (x^2-9)\,\,\,\, : \,\, |x|>3[/tex]
    [tex]2x (x+2)^2 \ge (x+2) (x^2-9)\,\,\,\, : \,\, |x|<3[/tex]

    Now we've got a cubic, but one factor is already out, so rearrange without re-absorbing the "outed" factor.

    [tex] (x^2 + 4x + 9)(x+2) \le 0\,\,\,\, : \,\, |x|>3[/tex]
    [tex] (x^2 + 4x + 9)(x+2) \ge 0\,\,\,\, : \,\, |x|<3[/tex]

    Here we have a cubic with just one real root at x=-2, so it's pretty easy to deduce where it's greater or less than zero. Just combine those regions (intersection) with the |x|>3 and |x|<3 constraints and make sure you omit the singular points that weren't in the original domain (-2).
     
  6. Sep 10, 2011 #5
    Re: Solve for x-- the inequality of quadratic

    Thank you very much.
     
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