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# x^2 - x < 0

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x^2 - x < 0 [#permalink]  13 May 2012, 12:54
x^2 - x < 0.

Why is this equal to 0<x<1?

When we take x as common:

x (x-1) < 0

Why is not x < 0 then?
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Re: x^2 - x < 0 [#permalink]  13 May 2012, 13:07
aazhar wrote:
x^2 - x < 0.

Why is this equal to 0<x<1?

When we take x as common:

x (x-1) < 0

Why is not x < 0 then?

u have already done the factorization : x (x-1) < 0

now assume that there is an '=' sign and get the values of x

u get two values : x=0 and x=1

Plot these on a number line and divide the number line on the inflection points

Number line will be divided into 3 parts

label the parts alternately as + ,-,+ (this can go on if u have more regions)
-- ensure that you start labelling from the right side and always start with a + and keep on alternating

as the inequality sign is negative .. u will choose the - region which is 0<x<1

If the inequality sign was positive

u would have chosen the 2 positive parts..

Hope this helps

Last edited by rni on 13 May 2012, 13:11, edited 1 time in total.
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Re: x^2 - x < 0 [#permalink]  13 May 2012, 13:10
This is a standard approach and can be used consistently for such questions
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Re: x^2 - x < 0 [#permalink]  14 May 2012, 09:38
aazhar wrote:
x^2 - x < 0.

Why is this equal to 0<x<1?

When we take x as common:

x (x-1) < 0

Why is not x < 0 then?

Given: a*b < 0 (i.e. ab is negative)
What can you say about the sign of a and b?

Can you say that 'a' MUST be negative? or that 'b' MUST be negative?
No! All you can say is that one and only one of them must be negative and the other must be positive ( -ve * +ve = -ve)

So how would you handle x(x-1) < 0?
Can you say x must be -ve? No!
Either x < 0 or (x-1) < 0 but not both.

Case 1: x < 0 and (x-1) > 0
x < 0 and x > 1
Is this possible? No! So no such values of x.

Case 2: x > 0 and (x - 1) < 0
x > 0 and x < 1
This gives us 0 < x < 1
So that is our solution.

Hope it makes sense now.
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Re: x^2 - x < 0   [#permalink] 14 May 2012, 09:38
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