Question

Absolute Value Question. Seriously need help.

Answer 1

\[[x / |x ^{2} -2|] - (1/5)\]

Answer 2

sorry additional
\[\ge 0\]

Answer 3

I am working on it

Answer 4

okay cool, thanks

Answer 5

is it?
\[\LARGE {x\over |x^2-2|}-\frac15\geq 0\]? (the question :) )

Answer 6

YESSS that is the question

Answer 7

lol , I'll "try " it too... although I don't promise anything :B haha..

Answer 8

... well:
\[ \LARGE {x\over |x^2-2|}-\frac15\geq 0 \]
\[ \LARGE {x\over |x^2-2|}\geq \frac15 \quad \quad |\cdot 5\]
\[ \LARGE {5x\over |x^2-2|}\geq 1 \quad \quad \cdot |x^2-2|\]
\[ \LARGE 5x\geq |x^2-2 |\]
I'm not sure something here.. Let me work on it. ...
also here's what wolframa says :)
http://www.wolframalpha.com/input/?i=%28x%2F%7Cx%5E2-2%7C%29-1%2F5%3E%3D0

Answer 9

before starting to solve it by quadratic equations:
let's see what the master tells us. I'm sure @satellite73 can handle this.

Answer 10

seems not to be here...
anyway , This is what I've got:
from:
\[\LARGE 5x\geq |x^2-2|\]
we have:
\[\LARGE x^2-2\leq 5x \quad ----[1]\] and
\[\LARGE x^2-2\geq -5x \quad -----[2]\]
[1]
\[\Large x^2-2-5x\leq 0\]
\[\Large x^2-5x-2\leq 0\]
now using the quadratic formula:
\[\LARGE {x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
I can't get Integers as wolframa said :( ...

Answer 11

yes, because the answers are in decimals

Answer 12

\[\LARGE {x_{1/2}} = \frac{{ - ( - 5) \pm \sqrt {25 - 4( - 2)} }}{2}\]
\[\LARGE {x_{1/2}} = \frac{{ 5 \pm \sqrt {33 } }}{2}\]
And here's the problem :( ...
really ?? so let's see ! ;)

Answer 13

\[\LARGE {x_1} = \frac{{5 - 5.74}}{2} = \frac{{ - 0.74}}{2} = - 0.37\]

Answer 14

what are multiple choices ?

Answer 15

sorry there's not
\[\LARGE x_1=-0.37\]
but
\[\LARGE x_1\leq -0.37\]

Answer 16

and for the second one [2]
we have:
\[\LARGE x^2-2\geq -5x\]
\[\LARGE x^2+5x-2\geq 0\]
using the quadratic formula we have:
\[\LARGE {x_{1/2}} \geq \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[\LARGE {x_{1/2}} \ge \frac{{ - 5 \pm \sqrt {33} }}{2}\]
\[\LARGE {x_{1/2}} \ge \frac{{ - 5 \pm 5.74 }}{2}\]
\[\LARGE {x_{1}} \ge \frac{{ - 5 - 5.74 }}{2}\]
\[\LARGE {x_{1}} \ge \frac{{ - 10 .74 }}{2}\]
\[\LARGE {x_{1}} \ge -5.37 \]

Answer 17

and x_2 ...

Answer 18

sdfsd
\[ \LARGE {x_{2}} \ge \frac{{ - 5 + 5.74 }}{2} \]
\[ \LARGE {x_{2}} \ge \frac{{ 0.74 }}{2} \]
\[ \LARGE {x_{2}} \ge 0.37 \]

Answer 19

sorry about "sdfsd" lol ... @fatinatikah still there? :)

Answer 20

yes yes im still here, im trying the solution u gave me :)

Answer 21

do you mind to post multiple choices? :)

Answer 22

@fatinatikah

Answer 23

\[\frac{x}{|x^2-2|}\geq\frac{1}{5}\]
\[5x\geq |x^2-2|\] so one thing we know for sure is that \(x\geq0\)because the right hand side is positive.
also we know if \(0<x<\sqrt{2},x^2-2<0\) and so \(|x^2-2|=2-x^2\)so on that interval you need to solve
\[5x>2-x^2\] get
\[x>\frac{-5+\sqrt{33}}{2}\]

Answer 24

if \(x\geq\sqrt{2}\) we need so solve
\[5x>x^2-2\] and get
\[x<\frac{5+\sqrt{33}}{2}\]

Answer 25

therefore you answer is
\[\frac{-5+\sqrt{33}}{2}\leq x\leq \frac{5+\sqrt{33}}{2}\]