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# How to solve |x|^2 - x >0

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Intern
Joined: 01 May 2015
Posts: 38
How to solve |x|^2 - x >0  [#permalink]

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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.

Answer: x > 1 or x < -1
Math Expert
Joined: 02 Aug 2009
Posts: 7764
How to solve |x|^2 - x >0  [#permalink]

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06 Jun 2016, 09:53
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sauravpaul wrote:
Can someone help me with this.

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

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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Posts: 38
Re: How to solve |x|^2 - x >0  [#permalink]

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06 Jun 2016, 19:29
Thanks chetan. Sorry I indeed mistyped the question. It is:

x^2 - |x| > 0

Math Expert
Joined: 02 Aug 2009
Posts: 7764
Re: How to solve |x|^2 - x >0  [#permalink]

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06 Jun 2016, 21:00
sauravpaul wrote:
Thanks chetan. Sorry I indeed mistyped the question. It is:

x^2 - |x| > 0

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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Posts: 38
Re: How to solve |x|^2 - x >0  [#permalink]

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06 Jun 2016, 22:16
3
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
Math Expert
Joined: 02 Aug 2009
Posts: 7764
Re: How to solve |x|^2 - x >0  [#permalink]

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06 Jun 2016, 22:34
sauravpaul wrote:
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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Intern
Joined: 01 May 2015
Posts: 38
Re: How to solve |x|^2 - x >0  [#permalink]

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07 Jun 2016, 04:49
Thank you very much for all your help.
Manager
Joined: 18 Feb 2018
Posts: 124
Re: How to solve |x|^2 - x >0  [#permalink]

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15 Jun 2019, 10:57
chetan2u wrote:
sauravpaul wrote:
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.

Re: How to solve |x|^2 - x >0   [#permalink] 15 Jun 2019, 10:57
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