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# M18-16

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Manager
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14 Apr 2013, 10:30
Could someone please explain how to get from the marked 1. to 2.?
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M18-16 (700).jpg [ 248.4 KiB | Viewed 976 times ]

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14 Apr 2013, 11:10
1
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x (x-1) (x+1) < 0

the solutions would require that of the given three factors (x, x-1, x+1) only if 1 or 3 of them are <0 then only the solution to x(x-1)(x+1) <0
i.e
x >0 and x+1>0 and x-1 <0
or x>0 and x-1>0 and x+1<0
or x+1>0 and x-1>0 and x< 0
or x<0 and x+1<0 and x-1<0

Now on solving

1: x >0 and x+1>0 and x-1 <0
we get x>0 and x>-1 and x<1
hence 0<x<1 is the solution range

2: x>0 and x-1>0 and x+1<0
we get x>0 and x>1 and x<-1
no possible solution range

3. x+1>0 and x-1>0 and x< 0
we get x>-1 and x>1 and x<0
no possible solution range

4. x<0 and x+1<0 and x-1<0
we get x<0 and x<-1 and x<1
hence the solution range is x<-1

therefore the solutions to x(x+1) (x-1) <0
are 0<x<1 or x<-1

another way to solve is to use the no. line
x(x-1)(x+1)<0

the solution of x(x-1) (x+1)=0 are x=-1,0,1

plotting these on no. line we get 4 regions

x<-1 , -1<x<0, 0<x<1 , and x>1

now for given exp: x(x-1) (x+1) we substitute for x values within the given range and check if it satisfies our criteria

if x<-1 exp <0

if -1<x<0 exp >0

if 0<x<1 exp <0

if x>1 exp >0

so for x(x-1) (x+1) <0
we need
x<-1 or 0<x<1
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15 Apr 2013, 00:29
1
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Expert's post
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14 Apr 2013, 11:48
$$x(x-1)(x+1)<0$$

we have to study the sign of each term.
1)the first one is -ve if x<0 2) the second is -ve if x<1 3)the third is -ve if x<-1
Write down on a line those numbers and the sign that the equation has.
now you have to intersect those interval and find the overall sign of the equation.
for instance x<-1: all are negative s$$o -*-*-=-$$ so in this interval its negative.
for -1<x<0 : $$-*+*-=+$$ so in this interval is positive
and so on... take a look at the image

Let me know if it's clear
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neg.png [ 2.03 KiB | Viewed 953 times ]

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19 Apr 2013, 07:12
HumptyDumpty wrote:
Could someone please explain how to get from the marked 1. to 2.?

Nothing new to add. Just another way of arriving at the solution.

x(x-1)(x+1)<0.

A product of three factors can only be negative if

1. All three of them are negative --> x<0 , x<1 , x<-1. Now, if x<-1, then automatically, the other two conditions are also fulfilled. Thus, one solution is at x<-1.

2.Any two of the factors are positive and only one is negative --> Say x>0 AND x>1 AND x<-1. Is this possible simultaneously? NO. Now consider x>1 AND x>-1 AND x<0. Again not possible. Finally consider x>0 AND x>-1 AND x<1. This is possible if x>0 AND x<1. Thus, the second solution is for 0<x<1.

But again this method is only for clearing the concept involved. The normal method to these problems is discussed above.
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Re: M18-16   [#permalink] 19 Apr 2013, 07:12
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# M18-16

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