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Thread: Inequalities with a modulus

  1. #1
    cpj
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    Inequalities with a modulus

    hi, how can i work out the values of x for this inequality please, |x/(x-2)| > |1/(x+1)|

    the most confusing thing is that both parts are being divided.
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  2. #2
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    Hello cpj
    Quote Originally Posted by cpj View Post
    hi, how can i work out the values of x for this inequality please, |x/(x-2)| > |1/(x+1)|

    the most confusing thing is that both parts are being divided.
    This is an awkward, 'bitty' sort of question. You need to look first at the values of $\displaystyle x$ where the sign of each expression changes, and then to consider the functions in the ranges formed by these values.

    So the signs change when $\displaystyle x = -1, 0, 2$. We consider, in turn, the four ranges $\displaystyle x<-1,\; -1<x<0,\;0<x<2,\; 2 < x$. I'll do the first two, and leave you to look at the last two.

    When $\displaystyle x < -1$
    $\displaystyle \frac{x}{x-2}>0\Rightarrow \left|\frac{x}{x-2}\right|=\frac{x}{x-2}$
    and
    $\displaystyle \frac{1}{x+1}<0\Rightarrow \left|\frac{1}{x+1}\right|=-\frac{1}{x+1}$
    So
    $\displaystyle \left|\frac{x}{x-2}\right|>\left|\frac{1}{x+1}\right|$

    $\displaystyle \Rightarrow \frac{x}{x-2}>-\frac{1}{x+1}$
    I'll leave you to check that this can be reduced to
    $\displaystyle \frac{x^2+2x-2}{(x-2)(x+1)}>0$
    Completing the square
    $\displaystyle \Rightarrow \frac{(x+1)^2-3}{(x-2)(x+1)}>0$
    Now for $\displaystyle x<-1, (x-2)(x+1) > 0$. So we want values of $\displaystyle x, \;(<-1)$ for which $\displaystyle (x+1)^2 > 3$

    $\displaystyle \Rightarrow |x+1| > \sqrt3$

    $\displaystyle \Rightarrow -(x+1) >\sqrt3$, since $\displaystyle x+1 < 0$

    $\displaystyle \Rightarrow x < -1-\sqrt3$

    So there's part of the solution set. (I said it was awkward!)

    Now we look at the next range, $\displaystyle -1<x<0$.

    In a similar way, we get this time:
    $\displaystyle \frac{x}{x-2}>\frac{1}{x+1}$
    which can be reduced to
    $\displaystyle \frac{x^2+2}{(x-2)(x+1)}>0$
    This time the denominator is negative, so we want values of $\displaystyle x$ for which
    $\displaystyle x^2+2<0$
    and clearly there are none. So there are no values of $\displaystyle x$ in the range $\displaystyle -1<x<0$.

    I'll leave you to look at the last two ranges of $\displaystyle x$ in the same way. I reckon the final answer is $\displaystyle \Big(-\infty, (-1-\sqrt3)\Big) \cup \Big((-1+\sqrt3), +\infty\Big)$.

    Grandad
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  3. #3
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    Thanks, Grandad (I had a similar problem).
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