amorfide
  • amorfide
Soooooo I need help with a modulus/absolute value equation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amorfide
  • amorfide
|dw:1350320124515:dw|
amorfide
  • amorfide
any method just explain
amorfide
  • amorfide
@ParthKohli

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