x^3 > 1 to x > 1

[asharma8080]

Is x^3 > 1?

(1) x > -2

(2) 2x - (b - c) < c - (b - 2)

I am sort of stuck at the fact that we can simplify x^3 > 1 to x > 1. Is there a better example that makes it more obvious that we can do this simplification?

For some reason, I do not see that. The explanation says that the cube does not "hide the sign."

I understand x^2 > 1 as x > 1 and x < -1

But, for some reason, I am not able to drill x^3 > 1 as x > 1 in my head.

Though, I am typing this, I am realizing that it is common sense that x will be greater than 1 if the cube is greater than 1. If x was negative, and we take its cube then it will be smaller than 1.

So, x^3 > 1 means x > 1 because x CAN NOT BE something like -2, -3, or even -(1/2), basically not less than 1.

[tim]

"hide the sign"; i'm not sure i like that wording either. :) so here's the thing: when you take an even root of both sides of an equation, you have to worry about the possibility of positive or negative roots. not so with odd roots. so you can actually take the cube root of both sides of an inequality with no negative consequences. it sounds like you've verified that this works with a few sample numbers, so that's good as it will solidify your intuitive understanding of the rule..

[geezer0305]

what is the OA?

i think its B.

[jnelson0612]

what is the OA?

i think its B.

Yes, correct!

[nocheivyirene]

My solution:

(1) x > -2
x= -1, (-1)^3 = -1 No!
x= 4, (4)^3 = 64 Yes!
INSUFFICIENT!

(2) 2x -(b-c) < c - b + 2
2x < c - c -b + b + 2
2x < 2
x < 1
x = 1/2, (1/2)^3 = 1/8 No!
x = 0, (0) No!
x= -1 N0!
SUFFICIENT!


Answer: B

[jlucero]

My solution:

(1) x > -2
x= -1, (-1)^3 = -1 No!
x= 4, (4)^3 = 64 Yes!
INSUFFICIENT!

(2) 2x -(b-c) < c - b + 2
2x < c - c -b + b + 2
2x < 2
x < 1
x = 1/2, (1/2)^3 = 1/8 No!
x = 0, (0) No!
x= -1 N0!
SUFFICIENT!


Answer: B

Absolutely correct. Do note, though, that if you can simplify the original equation: x^3 > 1? = x > 1?, you'll require less number picking here.

[nocheivyirene]

Is x^3 > 1?

(1) x > -2
(2) 2x - (b - c) < c - (b - 2)

My approach:
Whenever I encounter DS, I always try to quickly figure some possible values to the question before tackling statement (1) and statement (2).

x^3 > 1?
x=2 Yes
x=1 No
x=0 No
x=-1 No, If negative, x^3 is negative. All No for (-) values.

Statement 1: x > -2
Using my chart above, I know this is INSUFFICIENT.

Statement 2: 2x - (b - c) < c - (b - 2)
This is obviously some variable garbage that you must get rid of.
2x -b + c -c + b < 2
2x < 2
x < 1, Using my chart, it is always No. SUFFICIENT

Answer: B

[tim]

:)

[josesav7]

Hello,

Im not really clear on why B is correct. I did the same calculation and got 2x<2 or X<1. But then I stopped there because I did not know the sign of X, because if X were neg, then it could be X>-1, which is insufficient.

What am I missing? Maybe I didn't get the concepts right? Please help.

[tim]

It looks like you went from x<1 to x>-1, which is not a valid step under any circumstances. Remember, any time you get from one step to another, you need a valid mathematical operation that you've done to both sides. Please let me know if I've misinterpreted what you've done here, or if you have any further questions.

[josesav7]

Hi Tim - thanks for getting back to me.
I went from X<1 to X>-1, because in my head I thought that what if x was negative, then the signs would flip to X>-1. So in this way I thought that there could be two solutions depending on the sign of X. Because what I was thinking was that we cannout divide, unless we know the sign?

Are you saying doing that is incorrect?

[tim]

You don't flip a sign just because you know some information. You flip a sign as a direct result of doing something - multiplying or dividing by a negative. Since you haven't done that here, you wouldn't just flip a sign.