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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5

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Math Expert V
Joined: 02 Aug 2009
Posts: 8346
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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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.

--------(-1)---------(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!

Hi,
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..

example--
$$(x-3)^2*(x-2)>0$$..
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+1)(|x|-1)>0.
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.

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Posts: 1
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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Tips: The key to unlock confusion with “and” and “or” in inequality:

- Drawing graphs and crossing off all irrelevant values.
- “and” means must; “or” means could

Is x > 1?

(1) (x+1) (|x| - 1) > 0

(2) |x| < 5

(1): There are 2 cases: case 1 or case 2

Case 1: x+1 > 0 AND |x|-1 >0
X+1 >0 → x > -1 (A)
|x|-1 >0 → |x| >1 → x>1 OR x <-1 (B)

→ graph for A: //////////(-1)-------------(1)--------
→ graph for B: ---------(-1)///////////(1)--------

////////////means the values crossed off
----------------- means the values are accepted

Combining the two graphs above: this is the AND case, so we must cross off all values that don’t fit in any graph, represented by “//////////”
So the value of the 2 inequalities of case 1 is: x>1.

Case 2: x+1 <0 AND |x|-1 <0

X+1 < 0 → x <-1 (C)
|x|-1 <0 → |x| < 1 → -1<x<1 (D)

→ graph for C: ---------(-1)/////////(1)///////
→ graph for D: /////////(-1)----------(1)////////

As we see, there is no value of x that satisfy the both graphs. That means there is no solution for the inequalities in case 2.

→ So only case 1 is appropriate to consider. That means (x+1) (|x| - 1) > 0 when x>1. Sufficient.

(2) |x| < 5 → -5<x<5 → graph: //////-5--------1--------5//////

Clearly, x can be bigger than 1 or less than 1. Insufficient.

P/s: If you are still confused, you can test numbers!
_________________
“Knowing yourself is the beginning of all wisdom.” ―Aristotle.
Intern  S
Joined: 22 Apr 2018
Posts: 32
Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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Bunuel wrote:
Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;

Hi Bunuel,

can you please explain the red colored part, I got that x^2>1 and thus x will be either greater than 1 or less than -1 but how come we are only considering the part x>1? Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5   [#permalink] 30 Apr 2019, 11:18

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