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Solve for x in terms of a, the inequality

  1. Jan 6, 2006 #1
    Solve for x in terms of a, the inequality:
    \mid x^2 - 3ax + 2a^2 \mid < \mid x^2 + 3ax - a^2 \mid
    where [itex] x \in \mathbb{R}, a \in \mathbb{R}, a \neq 0[/itex]

    Squaring both sides, I get
    [tex]x^4 - 6ax^3 + 13a^2 x^2 - 12a^3 x + 4a^4 < x^4 + 6ax^3 + 7a^2 x^2 - 6a^3 x + a^4[/tex]

    [tex]12ax^3 - 6a^2 x^2 + 6a^3 x - 3a^4 > 0[/tex]

    [tex]4 ax^3 - 2a^2 x^2 + 2a^3 x - a^4 > 0[/tex]

    Stuck here. How do I proceed?
    Last edited: Jan 6, 2006
  2. jcsd
  3. Jan 6, 2006 #2


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    I think the best way to deal with a non-linear inequality is to start by solving the corresponding equation: in this case [itex]|x^2-3ax+ 2a^2|= |x^2+ 3ax- a^2|[/itex]. Yes, you can get rid of the absolute value by squaring both sides but that introduces higher powers. There are three possible cases:
    1) The quantities inside the absolute values on both sides are non-negative. In that case, the two absolute values are irrelevant: [itex]x^2- 3ax+ 2a^2= x^2+ 3ax- a^2[/itex] which is easy to solve.
    2) The quantities inside the absolute values on both sides are negative. In that case, the absolute values are just multiplied by -1. Divide both sides by -1 and you have the same equation as in 1.
    3) The quantities inside the absolute values on the two sides have opposite signs. In that case, again since we can multiply or divide by -1, it doesn't matter which side is negative: we have [itex]x^2- 3ax+ 2a^2= -(x^2+ 3ax- a^2)= -x^2- 3ax+ a^2[/itex] which just gives the quadratic equation [itex]2x^2= 3a^2[/itex], also easy to solve.

    The point is that the values of x where the two sides are equal separate the intervals on which one side is larger than the other. Choose one value of x in each interval to see whether the inequality is true in that case.
  4. Jan 6, 2006 #3


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    When squaring both sides, you seem to be missing a 9a2x2 on both sides, but it's okay, since they will cancel each other out.
    [tex]4 ax ^ 3 - 2a ^ 2 x ^ 2 + 2a ^ 3 x - a ^ 4 > 0[/tex].
    This is so far okay. To continue, you should factor it:
    [tex]4 ax ^ 3 - 2a ^ 2 x ^ 2 + 2a ^ 3 x - a ^ 4 > 0[/tex]
    [tex]\Leftrightarrow 2a x ^ 2 (2x - a) + a ^ 3 (2x - a) > 0[/tex]
    [tex]\Leftrightarrow (2x - a) (2a x ^ 2 + a ^ 3) > 0[/tex]
    [tex]\Leftrightarrow a (2x - a) (2x ^ 2 + a ^ 2) > 0[/tex]
    Note that 2x2 + a2 is already non-negative, so can you go from here?
  5. Jan 6, 2006 #4
    Great, thanks for the help!
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