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x^2 > x

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x^2 > x

by adthedaddy » Sun Aug 26, 2012 3:59 am
If x2 > x then how is x < 0 a solution?

I read in a solution that if x^2 > x then two solutions are possible --> x < 0 or x > 1
In my opinion, if x^2 > x then x^2 - x > 0 => x(x-1)>0 => Solution is x>0 or x>1

Please correct me if I am wrong and how does the book claim that x<0 as solution to the above.

(Source: MGMAT Book-3; Chap. Inequalities (Advanced))
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by neelgandham » Sun Aug 26, 2012 5:32 am
If x2 > x then how is x < 0 a solution?
x < 0. So, x is a negative number and x^2 is always a positive number. Since, a positive number is always greater than a negative number, x^2 is always greater than x
For e.g If x = -2, x^2 = 4. x^2 > x.
if x = -1/2, x^2 = 1/4. x^2 > x.
I read in a solution that if x^2 > x then two solutions are possible --> x < 0 or x > 1
Correct!
In my opinion, if x^2 > x then x^2 - x > 0 => x(x-1)>0 => Solution is x>0 or x>1 0
x>0 is not a solution. Why ? Square of any number between 0 and 1 is less then the number
If x = 1/2, x^2 = 1/4 and 1/4 < 1/2. i.e. x^2 < x
x>1 is part of the solution. Why ? Square of any number greater than 1 is greater than the number.
If x = 2, x^2 = 4 and 4>2/ i.e. x^2 > x.

Let me know if you have any further questions.

p.s: Even if the solution set of an in-equation is x>0 and x>1, we should denote it as x>0 as x>1 is a subset of x>0.
Anil Gandham
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by premnath » Sun Aug 26, 2012 8:01 am
x^2>x
=> x(x-1)>0
Hence,
Either both x and (x-1) are >0 this gives x>1
or both x and (x-1) are <0 this gives x<0

So, the equation is valid for all values where x<0 or x>1
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