Is x<0?

[VJesus12]

Is x<0?

1) x^5 < 0.
2) x^5 + x + 1 = 0.

The OA is the option D.

How can I get an answer from the second statement? Please, I need a clarification here. <i class="em em-cry"></i>

[Vincen]

Hello Vjesus12.

Let's take a look at your question.

We have to say if x<0 or not.

First statement

1) x^5 < 0.

Since the power is an odd number, then it implies that x will also have the same sign, so x<0. Therefore, this statement is sufficient.

Second statement

2) x^5 + x + 1 = 0.

This equation will lead us to $$x^5+x+1=0$$ $$x^5+x=-1$$ $$x\left(x^4+1\right)=-1 $$ $$x\left(x^4+1\right)=negative\ number .$$ Now, x^4+1 is always positive, so it implies that x must be negative, that is to say, x<0.

Therefore, this statement is sufficient.

In conclusion, each statement alone is sufficient. The answer is the option D.

I hope it helps you. <i class="em em-smiley"></i>

[Brent@GMATPrepNow]

Is x<0?

1) x^5 < 0.
2) x^5 + x + 1 = 0.

Target question: Is x<0?

Key Concept: ODD exponents preserve the sign of the base.
In other words, POSITIVE^ODD = POSITIVE and NEGATIVE^ODD = NEGATIVE

Statement 1: x^5 < 0
This is telling us that x^5 = NEGATIVE
So, it must be the case that x is NEGATIVE
So, the answer to the target question is YES, it is definitely the case that x < 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x^5 + x + 1 = 0
Subtract 1 from both sides to get: x^5 + x = -1
Applying the above concept, we know that x^5 and x share the same sign.
That is, EITHER x^5 and x are both positive, OR x^5 and x are both negative
However, in order for the equation x^5 + x = -1 to hold true, it must be the case that x^5 and x are both negative
So, the answer to the target question is YES, it is definitely the case that x < 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

[fskilnik@GMATH]

Is x<0?

1) x^5 < 0.
2) x^5 + x + 1 = 0.

How can I get an answer from the second statement? Please, I need a clarification here. <i class="em em-cry"></i>

Hi VJesus12,

x is a real number, therefore the trichotomy law applies to it: x is positive, x is zero or x is negative, and only one of these 3 possibilities occurs.

Now let´s consider statement (2) :

x is not zero (because 0^5 + 0 +1 would be equal to 1, not 0)
x is not positive (otherwise x^5 + x +1 would be greater than 1, not 0) , hence...
x is negative!

The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

[Jeff@TargetTestPrep]

Is x<0?

1) x^5 < 0.
2) x^5 + x + 1 = 0.

Statement One Alone:

x^5 < 0

When a negative number is raised to an odd integer exponent, the result is negative. Thus, the only way for x^5 to be less than 0 is if x is less than 0. Statement one is sufficient.

Statement Two Alone:

x^5 + x + 1 = 0

Let's re-express the given equation as x^5 + x = x(x^4 + 1) = -1. Since x^4 + 1 is nonnegative, that the only way for this to be true is if x is a negative number. Statement two is also sufficient.

Answer: D