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by rajanjindal27 » Thu Oct 06, 2011 12:05 am
-5/4>= x >= 5/2
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by Marysogh » Thu Oct 06, 2011 12:40 am
-5/2>= x >= 5/2
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by shankar.ashwin » Thu Oct 06, 2011 1:21 am
x-3|x| > +5 (Removing Mod, you get +/_5)

x + 3x > +5

So,

x + 3x > +5 (or) x-3x > +5

4x> +5 (or) 2x< +5

From here you could draw a number line and find the intersecting section of both is -5/4> x > 5/2
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by sl750 » Thu Oct 06, 2011 1:50 am
shankar.ashwin wrote:x-3|x| > +5 (Removing Mod, you get +/_5)

x + 3x > +5

So,

x + 3x > +5 (or) x-3x > +5

4x> +5 (or) 2x< +5

From here you could draw a number line and find the intersecting section of both is -5/4> x > 5/2
This inequality suggests that x is greater than or equal to 2.5 but less than or equal to -1.25. Shouldn't it be x>=5/2 or x<=-5/4 in one case and x<=-5/2 or x >= 5/4 ?
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by sanju09 » Thu Oct 06, 2011 2:44 am
seema19 wrote:How to identify the limits of this inequality?

|x - 3|x|| >= 5

[Mod x minus 3 mod x greater than or equal to 5.]
Basic learning is that if |n| ≥ 5, then there are only two possibilities viz. either n ≥ 5 or n ≤ -5. Therefore in the present case,

EITHER

x - 3|x| ≥ 5 → 3|x| ≤ x - 5

→ either 3 x ≤ x - 5 → x ≤ -5/2 or 3 x ≥ -x + 5 → x ≥ 5/4.

OR

x - 3|x| ≤ -5 → 3|x| ≥ x + 5

→ either 3 x ≥ x + 5 → x ≥ 5/2 or 3 x ≤ -x - 5 → x ≤ -5/4.

We may now safely pick x ≥ 5/2 and x ≤ -5/4 as the intervals that best explain the Absolute Value Inequality |x - 3|x|| ≥ 5.

Note: Please read → as "implies that". Formatting problem!
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