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Math Help - Modulus

  1. #1
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    Modulus

    Solve 4|x-a|=\frac{1}{x-a}
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  2. #2
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    Re: Modulus

    hmmm....
    say x is 4, and a is 3.5.... substitude them...... 4*0.5=2, 1/0.5=2
    as long as the difference is 0.5, its correct.....
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  3. #3
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    Re: Modulus

    Quote Originally Posted by Punch View Post
    Solve 4|x-a|=\frac{1}{x-a}
    Assuming that \displaystyle a > 0

    I suggest starting by noting that \displaystyle |x - a| = \begin{cases}x - a\textrm{ if }x \geq a\\ a - x\textrm{ if }x < a\end{cases}.

    So you need to look at two cases, Case 1:

    \displaystyle x \geq a \implies |x - a| = x - a, then

    \displaystyle \begin{align*}4|x - a| &= \frac{1}{x - a} \\ 4(x - a) &= \frac{1}{x - a} \\ 4(x - a)^2 &= 1 \\ (x - a)^2 &= \frac{1}{4} \\ x - a &= \pm \frac{1}{2} \\ x &= a \pm \frac{1}{2} \\ x &= \frac{2a \pm 1}{2} \end{align*}


    Case 2: \displaystyle x < a \implies |x - a| = a - x, then

    \displaystyle \begin{align*} 4(a - x) &= \frac{1}{x - a} \\ -4(x - a)^2 &= 1 \\ (x - a)^2 &= -\frac{1}{4} \\ x - a &= \pm \frac{i}{2} \\ x &= a \pm \frac{i}{2} \\ x &= \frac{2a \pm i}{2} \end{align*}


    So the solutions to the equation \displaystyle 4|x - a| = \frac{1}{x - a} are \displaystyle x = \left\{\frac{2a - 1}{2}, \frac{2a - i}{2}, \frac{2a + 1}{2}, \frac{2a + i}{2}\right\}
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