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Quadratic Absolute Value Inequality

  1. Sep 2, 2009 #1
    1. Solve:
    l2x^2+7x-15l<10



    3.

    I split it up into two cases and got:

    2x^2+7x-25<0 and 2x^2+7x-5>0

    With the quadratic formula I got:

    x= [-7+/- (249)^(1/2)]/4 and x= [-7+/- (89)^(1/2)]/4

    These clearly are the answers, but I am confused about the inequality from this point.
     
  2. jcsd
  3. Sep 2, 2009 #2

    VietDao29

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    Homework Helper



    Well, no, you don't split up in 2 cases like that:

    Here's a general way for you to solve equation of the type:
    |f(x)| < g(x)

    Since |f(x)| cannot take negative value (|f(x)| >= 0), and for the inequality (0 <=) |f(x)| < g(x) to hold (we are solving for x that makes the inequality holds), we must have g(x) > 0 (1).

    Note that if |A| < B, for some positive B, then -B < A < B.

    Applying that here, we have: -g(x) < f(x) < g(x). (2)

    From (1), and (2), we obtain:

    [tex]|f(x)| < g(x) \Rightarrow \left\{ \begin{array}{c} g(x) > 0 \\ -g(x) < f(x) < g(x) \end{array} \right.[/tex]

    Using the same reasoning, one can obtain:

    [tex]|f(x)| \leq g(x) \Rightarrow \left\{ \begin{array}{c} g(x) \geq 0 \\ -g(x) \leq f(x) \leq g(x) \end{array} \right.[/tex]

    In your problem, g(x) = 10 > 0, which is a constant, so, this should be easy.

    [tex]|2x ^ 2 + 7x - 15| < 10 \Rightarrow -10 < 2x ^ 2 + 7x - 15 < 10[/tex].

    Can you go from here? :)
     
    Last edited: Sep 2, 2009
  4. Sep 2, 2009 #3
    Thank you. However, this is one of the concepts that I should have learned in 8th grade, but never did. I followed your response completely, but I don't know where to go past
    5 < 2x^2 + 7x < 25

    Does it help to factor it originally into (x-1.5)(x+5) ?
     
  5. Sep 2, 2009 #4
    Now just split them

    [tex]5<2x^2+7x[/tex]

    &&

    [tex]2x^2+7x<25[/tex]
     
  6. Sep 2, 2009 #5
    Thanks.
     
  7. Sep 2, 2009 #6
    These lead to the original two cases the OP arrived at earlier.

    Quadratic inequalities (especially those whose roots are uncooperative radicals) can be solved by sign intervals.

    Take for example the inequality [itex]2x^2+7x-5>0[/itex]. The roots are

    [tex]r_1=\frac{-7-\sqrt{89}}{4},\;r_2=\frac{-7+\sqrt{89}}{4}. \text{ (As was mentioned. Note }r_1<r_2.)[/tex]

    The polynomial on the left factors into [itex]2(x-r_1)(x-r_2)[/itex] and this product will only be positive if both factors on the left are positive (i.e. [itex]x > r_1 \;\&\; x>r_2[/itex]) or both negative (i.e. [itex]x<r_1 \;\&\; x<r_2[/itex].) A potentially easier way of doing this is to realize that r1 and r2 subdivide the real line into three intervals in which the polynomial will have constant sign. Selecting a test value (even a nebulous one) from each and testing it into the polynmial will give the solution region:

    [tex]\begin{array}{cccc}\text{Interval} & (-\infty,r_1) & (r_1,r_2) & (r_2, \infty) \\
    \text{Test} & x<r_1 & r_1<x<r_2 & x>r_2 \\
    \text{Sign} & (+) & (-) & (+) \end{array}[/tex]

    We conclude that the solution region for this inequality is the intervals [itex](-\infty, r_1) \cup (r_2, \infty)[/itex].

    A similar approach to the other inequality will result in a solution region for it. The solution to the original absolute value inequality would be the intersection of the intervals found from the two cases.

    I hope this makes sence and is helpful.

    --Elucidus
     
  8. Sep 3, 2009 #7
    Thank you. I realized that it just led me to the same two cases I had obtained earlier. I discussed this problem with my dad, and we looked at it graphically. This led to basically the same solution as yours above.
     
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