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Soooooo I need help with a modulus/absolute value equation

Mathematics
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|dw:1350320124515:dw|
any method just explain

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Other answers:

|dw:1350320261447:dw| my graph
If |x| = y, then x = y OR x = -y
yes i get that part, so this gives |dw:1350321371149:dw|
then figure out x in both cases but i will get 4 answers, it only crosses 3 times, how do i know which one to eliminate?
|dw:1350321424876:dw| what you drew looks more exponential. The graph for y=|4/x| is linear with a jump at x=0.
There must be one for which x/2 + 3 is negative. That's an impossibility.
Because x/2 + 3 is the absolute value of something...?
|dw:1350321988874:dw| |dw:1350322071606:dw|
so what would my final answers be
Are you sure about the first part?
I think it should be -3 (+ or -) sqrt (21)
oh wait, sorry... my bad
in any case, -3-sqrt(17), try substituting that for x in x/2 + 3
And remember -3-sqrt(17) < -3-sqrt(16) = -7
thank you!
No problem... Sorry about my pathetic arithmetic :P
still made sense in a way :P
\[\rm \left|{4 \over x}\right| = {x \over 2} + 3 \qquad \Rightarrow \qquad {4 \over x} = \pm\left({x \over 2 }+3 \right)\]Let's solve both.\[\rm {4 \over x}={x \over 2}+3\]so\[8 = \rm x^2 + 6x\]therefore\[\rm x^2 + 6x-8=0 \]
For the other solution,\[{4 \over \rm x } = \rm -{x \over 2}-3\]so\[8=-\rm x^2 - 6x\]therefore\[\rm x^2 +6x +8=0\]
And as @zugzwang was kind enough to point out, make sure all values of x lead to \[\frac{x}{2}+3\] being nonnegative.

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