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Is x > 1?

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Re: Is x > 1? [#permalink]

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New post 12 Mar 2016, 04:47
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truongynhi wrote:
Hi Bunuel,

I have a question.

(x+1)(|x|-1)=0 has x=1 and x=-1 as its solution. However, x=-1 is a repeated root. Hence, on the number line, the sign will not change when it passes x=-1.


Therefore, only when x>1, (x+1)(|x|-1)>0.

Is my solution correct? I remember my highschool math teacher said something like 'the sign doesn't change when it passes a double root'. But, it's years ago and I just want to make sure that the approach is valid.

Thank you!

the solution here is correct, But I am not sure if it has to do with Double root in this specific instance..

Double root is generally when a polynomial has two equal roots..
so when ever you are placing the polynomial as>0, you cannot do anything to the squared part as it is always positive so it will depend on the other roots..

as can be seen \((x-3)^2\)will always be >0 except at x=3.. so we require to just look for x-2..
and x-2 will be positive only whenx>2..
so solution will be x>2 given that x is NOT equal to 3..

what is happening here --
x with a value <1 will make both (x+1) and (|x|-1) of oposite sign
a) x<-1
(x+1) is negtaive and (|x|-1) is positive
b) a) -1< x<1
(x+1) is positive and (|x|-1) is negative
so here (x+1)(|x|-1) will be negative and there will be no solutions for (x+1)(|x|-1)>0.


Absolute modulus :
Combination of similar and dissimilar things :

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Re: Is x > 1? [#permalink]

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New post 15 Apr 2018, 04:47
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Re: Is x > 1?   [#permalink] 15 Apr 2018, 04:47

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