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 < 11.
(x-1)^2<1(x - 1)^2 - 1^2 < 0(x - 1 + 1)(x - 1 - 1) < 0x(x - 2) < 00 < x < 22.
x^2-2<0x^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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