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Inequality problem

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Inequality problem [#permalink] New post 30 Oct 2004, 15:18
Solve the inequality : |x+2| > |3x-5|

What is the standard approach for these kinda problems?
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 [#permalink] New post 30 Oct 2004, 16:44
x+2 > (3x - 5) --> x<7/2
x+2 > -(3x-5) --> x>3/4
Hence, 3/4<x<7/2
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 [#permalink] New post 31 Oct 2004, 05:32
Paul wrote:
x+2 > (3x - 5) --> x<7/2
x+2 > -(3x-5) --> x>3/4
Hence, 3/4<x<7/2


Paul,

Don't we also have to consider 2 other cases:

1) -(x+2)> (3x-5)
2) -(x+2)> -(3x-5)

??
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 [#permalink] New post 31 Oct 2004, 07:20
I'd be happy if someone can give me a rule on this also. I actually considered the 2 cases you mentioned and I get the exact opposite answer to mine. I quickly tested it by plugging in numbers and invalidated it as wrong.
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 [#permalink] New post 03 Nov 2004, 19:13
|x+2| > |3x-5|


the check points are x=-2 and x=5/3

----------|-----------|-----------
-2 5/3

Case 1 : x <-2

-(x+2 ) > - (3x-5)
or x > 7/2 (not a good value)

Case 2 : x =-2

0 > -(3x-5) or x > 5/3 (not a good value)

Case 3 : -2<x<5/3

x+2 > -(3x-5)

x > 3/4

case 4 : x = 5/3

x+2 > 0 or x > - 2

case 5 : x > 5/3

x+2 > 3x -5

or x < 7/2



so combining all the above the solution is

3/4 < x < 5/3, x = 5/3 , 5/3< x < 7/2

or

3/4 < x < 7/2
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 [#permalink] New post 06 Nov 2004, 08:24
In case 3, shouldn't the negative sign be on the left-hand side rather than on the right-hand side?
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 [#permalink] New post 06 Nov 2004, 08:29
Quote:
Case 3 : -2<x<5/3

x+2 > -(3x-5)

x > 3/4


In this case x is greater than -2 , for example -1, so x+2 will always be positive. The other side will always be negative (3*-1 - 5 = -8). So the modulus to be positive the negative sign will be for 3x-5.

Let me know if this answers your question.
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 [#permalink] New post 06 Nov 2004, 08:57
The other approach when dealing with Mod on both sides of the inequality is to square both sides - one way to remove the mod. Mod is +ve and its square is +ve for sure too...thereby avoiding multiple cases to consider. This is a good rule when dealing with Mod when both sides have linear functions on both sides.

Looking at the problem under discussion, it would mean..

(x+2)^2 > (3x-5)^2

=> x^2 +4 + 4x > 9x^2 + 25 - 30x
=> -8x^2 + 34x - 21 > 0
=> (-2x+7)*(4x-3) > 0

=> x<7/2 and x> 3/4
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 [#permalink] New post 28 Dec 2004, 03:39
Venksune,

Thank you for reflecting this method.This is indeed an easier method to solve such problems.I have a quick question on the calculation:
x^2 +4 + 4x > 9x^2 + 25 - 30x
==> 8x^2 - 34x + 21 < 0
==> (2x-7)*(4x-3) < 0

Reflecting the above in a number line,I see the following:

<----------------------------------|
<----------|
0--------(3/4)-------------------(7/2)--------------


The intersecting section is when x<3/4. Thus would that not be the answer. But when I plug in 1/4, (2(1/4) - 7)*(4(1/4) -3) > 0
++++

I know I am missing something?Can you assist on this.Rgds,

Anna

venksune wrote:
The other approach when dealing with Mod on both sides of the inequality is to square both sides - one way to remove the mod. Mod is +ve and its square is +ve for sure too...thereby avoiding multiple cases to consider. This is a good rule when dealing with Mod when both sides have linear functions on both sides.

Looking at the problem under discussion, it would mean..

(x+2)^2 > (3x-5)^2

=> x^2 +4 + 4x > 9x^2 + 25 - 30x
=> -8x^2 + 34x - 21 > 0
=> (-2x+7)*(4x-3) > 0

=> x<7/2 and x> 3/4

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is there an easy way to factor [#permalink] New post 28 Dec 2004, 20:17
Hi, the only comment I would like to make is that it took me a real long time to factor the quadratic equation below. I used the quadratic formula. Is there an easier way to achieve this? If not, then this method may take longer then the one described earlier.

x^2 +4 + 4x > 9x^2 + 25 - 30x
==> 8x^2 - 34x + 21 < 0
==> (2x-7)*(4x-3) < 0
is there an easy way to factor   [#permalink] 28 Dec 2004, 20:17
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