Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5

Again you should consider 2*2 = 4 cases.

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

1. x < 0 --> (x+1)(-x - 1) > 0.

(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1. No solution here.

2. x < 0 --> (x+1)(-x - 1) > 0.

(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.

3. x > 0 --> (x+1)(x - 1) > 0.

(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1. This gives x > 1.

4. x > 0 --> (x+1)(x - 1) < 0.

(x + 1) < 0 and (x - 1) < 0 --> x < -1 and x < 1. No solution here.

So, (x+1)(|x| - 1) > 0 holds true only when x > 1.

Hope it's clear.

Hi Bunuel

Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.

\(-(x+1)^2>0\);

Add (x+1)^2 to both sides: \(0>(x+1)^2\), which is the same as \((x+1)^2<0\).

Is x > 1?

I would approach it the following way :

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

(+) * (+) > 0 or (-) * (-) > 0

For both of the parts to be positive we can see that x >1 . Just by trying few values you can figure this out. X cant be Zero as then the second part becomes - . X cant be 1 as then second part becomes 0 and hence the whole LHS becomes Zero.

For both of the parts to be negative we try any value less an Zero and see that no value will satisfy the equation. Hence X cannot be negative.. Hence A is Sufficient.

(2) |x| < 5

Clearly not sufficient

Answer is A.

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!

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)--------

Note about the graphs:
////////////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.

Answer: A.


P/s: If you are still confused, you can test numbers!

Bunuel KarishmaB
Can you please help me understand what mistake I am making in the attachment

Note that:

AB = 0 means that either A = 0 or B = 0 or both are 0
AB > 0 means that either Both are positive or Both are negative.
AB < 0 means that exactly one of them is positive and the other is negative.

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

means either (x + 1) > 0 AND (|x| - 1) > 0
or (x + 1) < 0 AND (|x| - 1) < 0

Now review your solution.

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