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Math Help - Modulus of function help

  1. #1
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    Modulus of function help

    Hello can anyone please tell me the steps in solving a modulus of a function?
    |2x +3| =5

    please i really need help
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  2. #2
    Super Member Quacky's Avatar
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    Re: Modulus of function help

    By definition, either 2x+3=5 or -(2x+3)=5. Alternatively, you could square both sides, to give (2x+3)^2=25 and solve as a quadratic.
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  3. #3
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    Re: Modulus of function help

    So if i get a question like this for exam all i have to do is square both sides???? AND THANKS ALOT>>>>>
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  4. #4
    Super Member Quacky's Avatar
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    Re: Modulus of function help

    It depends. Squaring both sides won't work for |2x+3|>5, for example. If you have an equation, it's a valid approach. I'd suggest knowing both methods, so that you can verify your answer.
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  5. #5
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    Re: Modulus of function help

    Quote Originally Posted by mathkid12 View Post
    Hello can anyone please tell me the steps in solving a modulus of a function?
    |2x +3| =5

    please i really need help
    You need to know the basic definitions of the modulus function, namely

    \displaystyle \begin{align*} |X| = \begin{cases} \phantom{-} X \textrm{ if }X \geq 0 \\ -X \textrm{ if } X < 0 \end{cases} \end{align*}

    and \displaystyle \begin{align*} \sqrt{X^2} = |X| \end{align*}.

    Also, by some geometric reasoning, you should know that \displaystyle \begin{align*} |X - c| = a \end{align*} means " \displaystyle \begin{align*} X \end{align*} is \displaystyle \begin{align*}  a \end{align*} units away from \displaystyle \begin{align*} c \end{align*}."


    So in the case of your equation \displaystyle \begin{align*} |2x + 3| = 5 \end{align*}, by the definition of the modulus function, that means

    \displaystyle \begin{align*} |2x + 3| &= \begin{cases} \phantom{-(} 2x + 3 \phantom{)} \textrm{ if }2x + 3 \geq 0 \\ -(2x + 3) \textrm{ if } 2x + 3 < 0 \end{cases} \\ &= \begin{cases} \phantom{-} 2x + 3 \textrm{ if } x \geq -\frac{3}{2} \\ -2x - 3 \textrm{ if }x < -\frac{3}{2} \end{cases} \end{align*}

    So now you need to solve the equation for the two separate cases.

    Case 1: \displaystyle \begin{align*} x \geq -\frac{3}{2} \end{align*}

    \displaystyle \begin{align*} |2x + 3| &= 5 \\ 2x + 3 &= 5 \\ 2x &= 2 \\ x &= 1 \end{align*}

    This solution is valid since it fits in with the restriction.


    Case 2: \displaystyle \begin{align*} x < -\frac{3}{2} \end{align*}

    \displaystyle \begin{align*} |2x + 3| &= 5 \\ -2x - 3 &= 5 \\ -2x &= 8 \\ x &= -4 \end{align*}

    This solution is also valid since it fits with the restriction.


    Now if you wanted to use the geometric interpretation...

    \displaystyle \begin{align*} |2x + 3| &= 5 \\ 2\left|x + \frac{3}{2}\right| &= 5 \\ \left|x + \frac{3}{2}\right| &= \frac{5}{2} \\ \left|x - \left(-\frac{3}{2}\right)\right| &= \frac{5}{2} \end{align*}

    So you are looking for the values of \displaystyle \begin{align*} x \end{align*} which are \displaystyle \begin{align*} \frac{5}{2} \end{align*} units away from \displaystyle \begin{align*} -\frac{3}{2} \end{align*}.

    Clearly these are \displaystyle \begin{align*} -\frac{3}{2} - \frac{5}{2} = -4 \end{align*} and \displaystyle \begin{align*} -\frac{3}{2} + \frac{5}{2} = 1 \end{align*}.
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  6. #6
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    Re: Modulus of function help

    wow thanks THANKS alot your help really appreciated
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