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Inequalities in modulus
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Author:  2iim [ Sun Oct 14, 2007 12:27 pm ]
Post subject:  Inequalities in modulus
For what range of values of x will the following inequality hold true?

|x + 7| > |2x + 3|
Author:  ag [ Sat Oct 20, 2007 9:03 am ]
Post subject:  should one look at all 4 cases
Should I be evaluating all four cases

case 1. when left hand side and right hand side are both positive

case 2: when left hand side is negative and right hand side is positive

case 3: reverse of case 2

case 4: when both sides are negative.

Do they ask such questions in CAT :!: Anyways, I will certainly skip this one. :wink:
Author:  2iim [ Thu Oct 25, 2007 11:21 am ]
Post subject:  Modulus on both sides, square both sides
AG

you do not have to evaluate all four cases. though that is one way to solve the question and as you mentioned a time consuming one.

it is sufficient if you square both sides of the inequation if you have modulus on both sides.

if |P| > |Q|, then P^2 will be greater than Q^2.

So, if you find range of values that satisfy P^2 > Q^2, then you have found out range of values that satisfy |P| > |Q|
Author:  ag [ Sat Oct 27, 2007 12:24 pm ]
Post subject: 
so, |x + 7| > |2x + 3| can be solved as (x + 7)^2 > (2x + 3)^2.

pls correct me if I have got it wrong

if i proceed with this, i get the following answer
(x + 7)^2 > (2x + 3)^2

x^2 + 14x + 49 > 4x^2 + 12x + 9
i.e., 3x^2 - 2x - 40 < 0
=> 3x^2 - 12x + 10x - 40 < 0
=> 3x (x - 4) + 10(x - 4) < 0
=> (3x + 10)(x - 4) < 0

so range of values will be -10/3 < x < 4

Is this correct?
Author:  2iim [ Mon Oct 29, 2007 8:13 pm ]
Post subject:  your answer is correct
ag - your answer is absolutely correct.

whenever there is a mod on both sides of the inequality as I had mentioned earlier all that you need to do is to square the values on both sides of the inequality and find the range of values.
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