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sauravpaul
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How to solve |x|^2 - x >0 06 Jun 2016, 09:41
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Can someone help me with this.

\(|x|^2 - x >0\)

I know we can substitute values and find. But I am looking for algebric solution.

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Answer: x > 1 or x < -1
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chetan2u
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How to solve |x|^2 - x >0 06 Jun 2016, 09:53
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sauravpaul
Can someone help me with this.

I know we can substitute values and find. But I am looking for algebric solution.

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Answer: x > 1 or x < -1


Hi,

you have not posted the Q correctly ..
But \(|x|^2-x>0\)...
or \(|x|^2>x\).....
It really does not make any difference that it is |x|^2 or x^2, as both will be POSITIVE and same value..

so \(x^2-x>0\)........................\(x(x-1)>0\)....................
so both x and x-1 should be of same sign...
1) If x>0, x-1>0....x>1..
2) if x<0, x-1<0....x<1..... since x<0 ans will be x<0...

so answer will be x>1 or x<0.......
You will have to recheck your answer x neednot be <-1 but x<0 will also satisfy..
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How to solve |x|^2 - x >0 06 Jun 2016, 19:29
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Thanks chetan. Sorry I indeed mistyped the question. It is:

x^2 - |x| > 0

Can you please help solve this.
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How to solve |x|^2 - x >0 06 Jun 2016, 21:00
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sauravpaul
Thanks chetan. Sorry I indeed mistyped the question. It is:

x^2 - |x| > 0

Can you please help solve this.


Hi,
\(x^2-|x|>0\)......
so \(x^2>|x|\).... since both sides are +ive, we can square two sides to eliminate MODULUS..
\(x^4>x^2.............x^4-x^2>0...............x^2(x^2-1)>0\)

ONLY one case is possible..
1) x^2 can NEVER be -ive, so x^2>0 and \(x^2-1>0..............(x-1)(x+1)>0\).....

This means that for all values of x between -1 and 1, x-1 will be -ive and x+1 will be +ive and thus their producy will be <0.........
and at x= 1 or x=-1, the equation will be equal to 0..
so our answer will be the range of x apart from mentioned above in BLUE colour....
x>1 and x<-1......
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How to solve |x|^2 - x >0 06 Jun 2016, 22:16
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Thanks. Can you advise if the following approach is ok:

|x| < x^2

This means that either:

Case 1: x < x^2 (when x is positive)

Or

Case 2: -x < x^2 (when x is negative)

Case 1: x < x^2 (when x is positive)
x^2 - x > 0
x(x-1) > 0

Solutions of this:
x > 1
OR
x < 0

But x < 0 is not possible, because this case is for when x is positive (is this reasoning correct?)

So, the only solution is: x > 1


Case 2: -x < x^2 (when x is negative)
x^2 + x > 0
x(x+1) >0

Solutions of this:
x >0
OR
x < -1

But x > 0 is not possible, because this case is for when x is negative (is this reasoning correct?)

So, the only solution is: x < -1
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How to solve |x|^2 - x >0 06 Jun 2016, 22:34
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sauravpaul
Thanks. Can you advise if the following approach is ok:

|x| < x^2

This means that either:

Case 1: x < x^2 (when x is positive)

Or

Case 2: -x < x^2 (when x is negative)

Case 1: x < x^2 (when x is positive)
x^2 - x > 0
x(x-1) > 0

Solutions of this:
x > 1
OR
x < 0

But x < 0 is not possible, because this case is for when x is positive (is this reasoning correct?)

So, the only solution is: x > 1


Case 2: -x < x^2 (when x is negative)
x^2 + x > 0
x(x+1) >0

Solutions of this:
x >0
OR
x < -1

But x > 0 is not possible, because this case is for when x is negative (is this reasoning correct?)

So, the only solution is: x < -1

Yes, you are correct....
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How to solve |x|^2 - x >0 07 Jun 2016, 04:49
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Thank you very much for all your help.
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How to solve |x|^2 - x >0 15 Jun 2019, 10:57
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chetan2u
sauravpaul
Thanks. Can you advise if the following approach is ok:

|x| < x^2

This means that either:

Case 1: x < x^2 (when x is positive)

Or

Case 2: -x < x^2 (when x is negative)

Case 1: x < x^2 (when x is positive)
x^2 - x > 0
x(x-1) > 0

Solutions of this:
x > 1
OR
x < 0

But x < 0 is not possible, because this case is for when x is positive (is this reasoning correct?)

So, the only solution is: x > 1


Case 2: -x < x^2 (when x is negative)
x^2 + x > 0
x(x+1) >0

Solutions of this:
x >0
OR
x < -1

But x > 0 is not possible, because this case is for when x is negative (is this reasoning correct?)

So, the only solution is: x < -1

Yes, you are correct....

In Case I, how is x>1 or x<0, shouldn't it be x>1 or x>0 and we consider only the higher limit. ie. x>1.
In Case II , how is x>0 or x<-1, shouldn't or be x>0 or x>-1.

Please help, I am really confused.
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How to solve |x|^2 - x >0 07 Apr 2023, 05:49
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