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# Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0

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Re: Is x negative? [#permalink]
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IMO C:

A) X^2(1-X^2) <0 so X^3<0 or X^2>1
x could be positive or negative

B) X^2-1<0
(X-1)(X+1)<0 so X<1 or X<-1
x could be positive or negative

Combined X is negative for both statements to hold true.

OA?
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Re: Is x negative? [#permalink]
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IMO C.

1, is equivalent to x(x-1)(x+1) >0 ( x^2 is always>0 and i have inverted the sign)

this gives x>1 and -1<x<0 , thus not sufficient

2, gives -1< x<1 not sufficient.

if you combine the result we get -1<x<0 , thus x is negative and thus C.

Another way, since we know both are not sufficient individually.

2nd equation gives x^2 -1 <0 => 1- x^2 >0
now in first equation, if its -ve then x^3 must be -ve....thus both taken together sufficient.
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Re: Is x negative? [#permalink]
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Alternative approach by taking numbers.
Statement1: $$x^3(1-x^2) < 0$$
From this we can say $$x <> 0$$
and $$x <>+1 or -1$$.

Consider x=2 Then the equation results in $$8*-3 <0$$. Satisfy the equation.
Consider x= 1/2 then the equation results in $$1/8*(3/4) >0$$. Does not satisfy the equation.

Consider x=-2 Then the equation results in $$-8*-3 >0$$. Does not satisfy the equation.
Consider x= -1/2 then the equation results in $$-1/8*(3/4) <0$$.Satisfy the equation.

This statement is not sufficient to conclude.

Statement2: $$x^2-1 <0$$
From the statement we can say that $$x <>=1 or -1$$
Consider x=2 then the equation results in $$3 >0$$. Does not satisfy the equation.
Consider x=1/2 then the equation results in $$-3/4 <0$$. Satisfy the equation.

Consider x=-2 then the equation results in $$3 >0$$. Does not satisfy the equation.
Consider x=-1/2 then the equation results in $$-3/4 <0$$. Satisfy the equation.

This statement alone is not sufficient.

Combining both the statements when x is -ve that is -1/2 , it satisfies both the equations.

Ans is C.
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Re: Is x negative? [#permalink]
1. x^3(1-x^2) <0
x^3- x^5<0
x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).

2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).

the answer is A
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Re: Is x negative? [#permalink]
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abhi758 wrote:
Is X negative?

1. $$x^3(1-x^2) < 0$$
2. $$x^2-1 < 0$$

Is $$x<0$$?

(1) $$x^3(1-x^2) < 0$$
Two cases:
A. $$x>0$$ and $$1-x^2<0$$, which means $$x<-1$$ or $$x>1$$ --> $$x>1$$;
B. $$x<0$$ and $$1-x^2>0$$, which means $$-1<x<1$$ --> $$-1<x<0$$.

So $$x^3(1-x^2) < 0$$ holds true for two ranges of $$x$$: $$x>1$$ and $$-1<x<0$$. If $$x$$ is in the first range answer to the question is NO, but if $$x$$ is in the second range answer to the question is YES. Two different answers. Not sufficient.

(2) $$x^2-1<0$$ --> $$-1<x<1$$. x can be positive as well as negative. Not sufficient.

(1)+(2) $$x>1$$ or $$-1<x<0$$ AND $$-1<x<1$$ --> intersection of the ranges from (1) and (2) is $$-1<x<0$$. $$x$$ is negative. Sufficient.

aiyedh wrote:
1. x^3(1-x^2) <0
x^3- x^5<0
x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).

2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).

the answer is A

Red part is not correct. $$x^3 < x^5$$ does not hold true for all positive $$x-es$$. For example $$x=0.5$$ --> $$x^3>x^5$$. Also $$x$$ can be negative in the range $$-1<x<0$$ and $$x^3 < x^5$$ will hold true.
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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Bunuel wrote:
Is x negative?

(1) $$x^3(1 - x^2) < 0$$

(2) $$x^2 - 1 < 0$$

My reasoning:

Statement 1: either $$x^3$$ is negative or $$(1 - x^2)$$ is negative. They are opposite signs. However we can have a positive x and a negative x in this case.

Testing values of 2 and $$\frac{-1}{3}$$ will prove this.

Statement 2: $$x^2 - 1 < 0$$ -> $$x^2 < 1$$ so $$x > -1 & x < 1$$ Again inconclusive because we can have any positive or negative fraction.

Combining the two statements only "negative fractions" are in our overlapping set. The answer is C. x MUST be less than 0
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Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
Hi chetan2u, Bunuel, Gladiator59

I am not able to figure out how we are able to find solution combining both the statement for above question, Could you please assist.
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Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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Gmatprep550 wrote:
Hi chetan2u, Bunuel, Gladiator59

I am not able to figure out how we are able to find solution combining both the statement for above question, Could you please assist.

The question is
Quote:
Is x negative?

(1) $$x^3(1 - x^2) < 0$$

(2) $$x^2 - 1 < 0$$

Let us take both combined..

Statement II tells us $$x^2 - 1 < 0$$, so when we multiply this with -1, the inequality sign will change
=>$$-1(x^2 - 1) > 0*(-1).....1-x^2>0$$ We did this to get the term of statement I..

Now take statement I..
$$x^3(1 - x^2) < 0$$ ... Here we see from statement II that $$1-x^2>0$$, so $$x^3(1 - x^2) < 0$$ => $$x^3(positive..value) < 0$$ , thus $$x^3<0...x<0$$
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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Is x negative?

(1) x^3(1−x^2)<0

(2) x^2−1<0

_________________

IMO, it is C.

The question is x < 0?
(1) divide both RHS and LHS by x^2.
then you have, x (1-x^2) <0.
now, take "-1" out from the parenthesis. -x(x^2 - 1) < 0.
Divide both RHS and LHS by -1.
then you have x (x^2 - 1) > 0.
In other words, x ( x+1) (x-1) >0.
so -1<x<0 and x>1
insufficient.

(2) we can say ( x+1) (x-1) <0
so -1<x<1
insufficient.

(1) + (2) together
-1<x<0.
so x is negative.
sufficient to answer whether x is negative.
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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Bunuel wrote:
Is x negative?

(1) $$x^3(1 - x^2) < 0$$

(2) $$x^2 - 1 < 0$$

Graphical solution:
Attachment:

Screenshot.png [ 151.5 KiB | Viewed 20340 times ]

1) $$x^3(1 - x^2) < 0$$
Case I: $$x^3 < 0$$ and $$(1 - x^2) > 0$$
x < 0 and $$x^2 < 1$$
x < 0 and - 1 < x < 1
So, - 1 < x < 0 YES

Case II: $$x^3 > 0$$ and $$(1 - x^2) < 0$$
x > 0 and $$x^2 > 1$$
x > 0 and x > 1 and x < -1
So, x > 1 YES

INSUFFICIENT.

2) $$x^2 - 1 < 0$$
$$x^2 < 1$$
- 1 < x < 1
YES and NO both cases.

INSUFFICIENT.

Together 1 and 2
Common range is -1 < x < 0

SUFFICIENT.

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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
Is x negative?
Is $$x < 0$$?

(1) $$x^3(1 - x^2) < 0$$

Either x^3 or (1 - x^2) is negative.

If $$x^3 < 0$$ then

$$1 - x^2 > 0$$
$$x^2 < 1$$
$$-1 < x < 1$$

If $$x^3 > 0$$ then
$$1 - x^2 < 0$$
$$x^2 - 1 > 0$$
$$(X+1)(X-1) > 0$$
$$X > 1$$ and $$x < -1$$

INSUFFICIENT.

(2) $$x^2 - 1 < 0$$
$$(x+1)(X-1) < 0$$
$$-1 < x < 1$$

INSUFFICIENT.

(1&2) Combined, the range is $$-1 < x < 0$$.

Thus, $$x < 0$$.

SUFFICIENT.
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
Could someone explain how can we transform x^2-1<0 to -1<x<1. I mean I understand it logically but algebraically we have (x+1)(x-1)<0 => x<-1 or x<1
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
2020prep2020 wrote:
Could someone explain how can we transform x^2-1<0 to -1<x<1. I mean I understand it logically but algebraically we have (x+1)(x-1)<0 => x<-1 or x<1

It's a matter of rearranging

x^2 - 1 < 0
x^2 < 1
\sqrt{x^2} < \sqrt{1}
| x | < 1 <-----MEMORIZE this...anytime you see the square root of a variable raised to an even exponent, you end up with the absolute value of that variable

Hence, -1 < x < 1.

This might be a silly question, but suppose after combining both statements we find that there are no areas of overlap. Then the answer would be E?
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
Is X < 0 ?

(S1) (x)^3 * (1 - (x)^2) < 0

Case 1: (x)^3 < 0 ------and------- 1 - (x)^2 > 0

X < 0 ------ and -------(x)^2 < 1

X < 0 ------and------- (-)1 < X < +1

Summary of Case 1: if X < 0 -------- (-)1 < X < 0

Case 2: (x)^3 > 0 ------ and -------- 1 - (x)^2 < 0

X > 0 ----- and ------ (X)^2 > 1

Summary of Case 2: if X > 0 ------- X > +1

X can be Negative or Positive

S1 NOT Sufficient alone

(2)
(x)^2 < 1

[X] < 1

-1 < X < +1

X can be Negative or Positive
S2 NOT Sufficient Alone

Together:

If X < 0:

Statement 1 says: -1 < X < 0
and
Statement 2 says: -1 < X < 0

If X > 0

Statement 1 says: X > +1
and
Statement 2 says: 0 < X < +1

if X is a Positive Value, it can NOT Satisfy S1 and S2 TOGETHER at the Same Time

therefore, X < 0

and Definite YES: X is Negative

(C) Together Sufficient
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]

Could you please explain why the below solution is incorrect?

aiyedh wrote:
1. x^3(1-x^2) <0
x^3- x^5<0
x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).

2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).

the answer is A
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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Laksh47 wrote:

Could you please explain why the below solution is incorrect?

aiyedh wrote:
1. x^3(1-x^2) <0
x^3- x^5<0
x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).

2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).

the answer is A

When x is between 0 and -1, say x is -1/2.
$$\frac{-1}{2}^3=\frac{-1}{8}$$
$$\frac{-1}{2}^5=\frac{-1}{32}$$
$$\frac{-1}{32}>\frac{-1}{8}$$

So x can be negative too.
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Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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Laksh47 wrote:

Could you please explain why the below solution is incorrect?

aiyedh wrote:
1. x^3(1-x^2) <0
x^3- x^5<0
x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).

2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).

the answer is A

$$x^3 < x^5$$

does not imply that x is positive. This is satisfied for x = 2 as well as x = -1/2.
Note that the number properties vary over 4 basic intervals of the number line

x < -1
-1 < x< 0
0 < x < 1
and x > 1

Also, -1, 0 and 1 behave differently. It is best to use

$$x^3 - x^5 < 0$$
$$x^5 - x^3 > 0$$
$$x^3 (x^2 - 1) > 0$$
$$x^3 (x + 1)(x - 1) > 0$$

Transition points are 0, -1, 1.
x will be positive when x > 1 or -1 < x < 0.
Re: Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0 [#permalink]
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