How to solve |x|^2 - x 0

Question

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

chetan2u · Answer

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..

sauravpaul · Answer

Thanks chetan. Sorry I indeed mistyped the question. It is:

x^2 - |x| > 0

Can you please help solve this.

chetan2u · Answer

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......

sauravpaul · Answer

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

chetan2u · Answer

Yes, you are correct....

sauravpaul · Answer

Thank you very much for all your help.

GittinGud · Answer

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

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