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# How to calculate range of x in inequality equations with MOD

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Manager
Joined: 08 Dec 2006
Posts: 79
Location: United States
Concentration: Entrepreneurship, General Management
Schools: Harvard Business School (HBS) - Class of 2014
GPA: 3.64
WE: Engineering (Aerospace and Defense)
Followers: 32

Kudos [?]: 74 [0], given: 3

How to calculate range of x in inequality equations with MOD [#permalink]  11 Jun 2011, 14:43
is |x+1|<2?

1. (x-1)^2<1
2. x^2-2<0

I need to understand reason behind why some approaches are not yielding the right range of x when i am trying to simplify the questions.

first approch:
[x+1]<2
(x+1)^2 <4
X^2+2x+1 <4
(x+3)(x-1)<0
X+3<0
x-1<0
x<-3
x <1
WRONG result

2)[x+1]<2
(x+1)^2 <4
X^2+2x+1 <4
X(x+2) <3
X<3
X<-1
WRONG RESULT

3)[x+1]<2
(x+1)^2<4
Taking root on both sides
(x+1)<2 ; (x+1)<-2
X<1; x<-3
WRONG RESULT

4)[x+1]<2
X+1 = y
|y|<2
-2<y<2
-2<x+1<2
-3<x<1
CORRRECT RESULT

5) [x+1]<2
X+1 <2
-(x+1)<2
X<1; x>-3

CORRECT Result

SO I need to understand why the first three approach are not giving me right range???what am I doing wrong. Please explain the underlying concept between modulus/squares/roots.

Thank you,

Regards,
ABhi
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Manager
Joined: 16 May 2011
Posts: 204
Concentration: Finance, Real Estate
GMAT Date: 12-27-2011
WE: Law (Law)
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Kudos [?]: 51 [0], given: 37

Re: How to calculate range of x in inequality equations with MOD [#permalink]  13 Jun 2011, 14:17
if i understand the Q properly than the question is:

is x lies between -3 and 1?

statement 1 says that x is less than 2 or 0 which concludes x<0. SO X CAN BE WITHIN THE RANGE(-3 to 0) or it cna be -4,-5 etc... hence insuf.

statement 2 says (and i hope i opened it correctly) that x is less than root of 2. hence x can be within the range (0, -1) or out (-5, -6 etc..)

combinining both of them still leave 2 options whithin t6he range or without.

hence tha answer E and god help me if i got it wrong
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Re: How to calculate range of x in inequality equations with MOD [#permalink]  13 Jun 2011, 19:17
3
KUDOS
Expert's post
Abhishek.pitti wrote:
is |x+1|<2?

1. (x-1)^2<1
2. x^2-2<0

I need to understand reason behind why some approaches are not yielding the right range of x when i am trying to simplify the questions.

first approch:
[x+1]<2
(x+1)^2 <4
X^2+2x+1 <4
(x+3)(x-1)<0
X+3<0
x-1<0
x<-3
x <1
WRONG result

2)[x+1]<2
(x+1)^2 <4
X^2+2x+1 <4
X(x+2) <3
X<3
X<-1
WRONG RESULT

3)[x+1]<2
(x+1)^2<4
Taking root on both sides
(x+1)<2 ; (x+1)<-2
X<1; x<-3
WRONG RESULT

4)[x+1]<2
X+1 = y
|y|<2
-2<y<2
-2<x+1<2
-3<x<1
CORRRECT RESULT

5) [x+1]<2
X+1 <2
-(x+1)<2
X<1; x>-3

CORRECT Result

SO I need to understand why the first three approach are not giving me right range???what am I doing wrong. Please explain the underlying concept between modulus/squares/roots.

Thank you,

Regards,
ABhi

When you want to find the range of numbers that satisfy an inequality, you CANNOT square both sides. You can square both sides only in case of an equation.

Why?
Say -4 < 2 is the inequality. You square both sides.
16 < 4 Is it true? No. The inequality may not hold after you square both sides.

This is how you solve inequalities:

$$|x+1|<2$$
$$-2 < x + 1 < 2$$
$$-3 < x < 1$$

1.
$$(x-1)^2<1$$
$$(x - 1)^2 - 1^2 < 0$$
$$(x - 1 + 1)(x - 1 - 1) < 0$$
$$x(x - 2) < 0$$
$$0 < x < 2$$

2.
$$x^2-2<0$$
$$x^2 - \sqrt{2}^2 < 0$$
$$(x - \sqrt{2})(x + \sqrt{2}) < 0$$
$$-\sqrt{2} < x < \sqrt{2}$$

Both together are insufficient.
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Re: How to calculate range of x in inequality equations with MOD [#permalink]  14 Jun 2011, 00:53
Very good explanation!
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Re: How to calculate range of x in inequality equations with MOD [#permalink]  11 Aug 2014, 20:48
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Re: How to calculate range of x in inequality equations with MOD   [#permalink] 11 Aug 2014, 20:48
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