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|1/(2+a)|<1, a?

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  1. Mar 10, 2015 #1
    Hi

    I'm trying to solve this inequality

    |1/(2+a)| < 1.

    1/(2+a) < 1 ∨ 1/(2+a) > -1

    1 < 2+a
    a > -1

    and

    1 > -2-a
    3 > -a
    a > -3

    I know that the boundaries are
    -∞ < a < -3 ∨ -1 < a < ∞

    What have I done wrong?

    thanks in advance
     
  2. jcsd
  3. Mar 10, 2015 #2

    symbolipoint

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    |1/(2+a)|-1<0, result of adding -1 to both sides.


    Examine two conditions separately. The case, 2+a>=0, and the case, 2+a<0. Follow each solution process separately.

    I can't think in advance if they will be conjoint or disjoint, but you'll find out once both conditions are solved.

    Be aware, the expression 2+a must not be allowed equal to zero, meaning a<>-2;
    so -2 is a critical value.
     
  4. Mar 10, 2015 #3

    Mark44

    Staff: Mentor

    The connector above should be ∧ ("and") rather than "or".

    If |x| < b, then an equivalent inequality is -b < x < b. This is what you should have done here.
    When you multiply both sides by 2 + a, are you multiplying by a positive number or a negative number? It makes a difference as regards the inequality symbol.
     
  5. Mar 10, 2015 #4

    mathman

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    The case with the negative sign should read 1/(2+a) < -1. Take absolute value on both sides is equivalent to changing sign on both sides, which requires reversing the inequality direction.
     
  6. Mar 11, 2015 #5

    statdad

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    For an alternate approach
    [tex]
    \begin{align*}
    -1 & < \frac 1 {2+a} < 1 \\
    -(2+a)^2 & < (2+a) < (2+a)^2 \\
    -a^2 - 4a - 4 & < 2+a < a^2 + 4a + 4 \\
    -a^2 - 5a - 6 & < 0 < a^2 + 3a + 2 \\
    -\left(a^2 + 5a + 6\right) < 0 < a^2 + 3a + 2 \\
    -(a+3)(a+2) < 0 < (a+1) (a+2)
    \end{align*}
    [/tex]
    Solve the right inequality, solve the left inequality. The solution is the intersection of the two parts.
     
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