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amorfide
 4 years ago
Soooooo I need help with a modulus/absolute value equation
amorfide
 4 years ago
Soooooo I need help with a modulus/absolute value equation

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amorfide
 4 years ago
Best ResponseYou've already chosen the best response.0any method just explain

amorfide
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350320261447:dw my graph

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

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

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

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

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

amorfide
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350321988874:dw dw:1350322071606:dw

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

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

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

zugzwang
 4 years ago
Best ResponseYou've already chosen the best response.1oh wait, sorry... my bad

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

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

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

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

ParthKohli
 4 years ago
Best 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\[\rm x^2 + 6x8=0 \]

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0For 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
 4 years ago
Best ResponseYou've already chosen the best response.0And 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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