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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