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amorfide
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Soooooo I need help with a modulus/absolute value equation
 2 years ago
 2 years ago
amorfide Group Title
Soooooo I need help with a modulus/absolute value equation
 2 years ago
 2 years ago

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amorfide Group TitleBest ResponseYou've already chosen the best response.0
dw:1350320124515:dw
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
any method just explain
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
@ParthKohli
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
dw:1350320261447:dw my graph
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
If x = y, then x = y OR x = y
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
yes i get that part, so this gives dw:1350321371149:dw
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

2le Group TitleBest ResponseYou've already chosen the best response.0
dw:1350321424876:dw what you drew looks more exponential. The graph for y=4/x is linear with a jump at x=0.
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
There must be one for which x/2 + 3 is negative. That's an impossibility.
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
Because x/2 + 3 is the absolute value of something...?
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
dw:1350321988874:dw dw:1350322071606:dw
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
so what would my final answers be
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
Are you sure about the first part?
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
I think it should be 3 (+ or ) sqrt (21)
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
oh wait, sorry... my bad
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
in any case, 3sqrt(17), try substituting that for x in x/2 + 3
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
And remember 3sqrt(17) < 3sqrt(16) = 7
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
thank you!
 2 years ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.1
No problem... Sorry about my pathetic arithmetic :P
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
still made sense in a way :P
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\[\rm \left{4 \over x}\right = {x \over 2} + 3 \qquad \Rightarrow \qquad {4 \over x} = \pm\left({x \over 2 }+3 \right)\]Let's solve both.\[\rm {4 \over x}={x \over 2}+3\]so\[8 = \rm x^2 + 6x\]therefore\[\rm x^2 + 6x8=0 \]
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
For the other solution,\[{4 \over \rm x } = \rm {x \over 2}3\]so\[8=\rm x^2  6x\]therefore\[\rm x^2 +6x +8=0\]
 2 years ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
And as @zugzwang was kind enough to point out, make sure all values of x lead to \[\frac{x}{2}+3\] being nonnegative.
 2 years ago
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