og ds 154

You need to get your basics right...
The question is already discussed in the forum...

You are trying number substitution for equations that do not require number substitution. Then you pick wrong set of numbers, and then you say you don't understand the logic....

The logic is simple, if the result of multiplication is negative, then either of the two numbers is negative, i.e. xy <0> either x < 0 or y < 0, but not both

ok i dont understand why you are choosing your tone of voice. You are not obligated to answer my questions. I am simply trying to understand a topic. If you do not like please refrain from my questions. You may be right that this topic may be in other parts of the forum. Dear friend- I am LEARNING and trying to LEARN.

Hi,
Look at eqn 2, and try and reverse it,
i.e. if x^2-1<0> 0 < 1 -x^2 (Step I)

which means thats (1 -x^2) is +ve in eqn 1.
Also. a product of +ve no and a -ve no will produce a -ve no i.e. a no less than 0.

Now look at eqn 1.
(x^3 ) (1-x^2)<0

x^3 must be -ve since (1-x^2) is positive.

x^3 is -ve implies that x is -ve.

i aM GOING TO REARRANGE IN ORDER TO GRASP THIS CONCEPT BUT i AM STILL A BIT LOST. SORRY ! WHAT DOES ve MEAN BY THE WAY?

Enginpasa1 wrote:154.

Is X negative?

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

(1) For this to be negative, exactly 1 of the two terms must be negative.

If we pick x = 5, the second term is negative, so the whole thing will be negative. Is 5 negative? No.

If we pick x = -(1/2), the first term is negative, so the whole thing will be negative. Is -(1/2) negative? Yes.

Therefore, (1) is insufficient.

(2) Let's simplify this to x^2 < 1.

We can pick either +(1/2) or -(1/2), so statement (2) is also insufficient.

When combining, let's start with the simpler statement. If x^2<1, then -1 < x < 1 (in other words, x is a negative fraction, 0 or a positive fraction).

Now let's look at statement (1)

x=0 isn't a solution to 1, so we can ignore it.

If x is a positive fraction, then both terms will be positive. Therefore, x can't be a positive fraction.

Since we've eliminated 0 and positive fractions, the only remaining possibilities are negative fractions. Therefore, x MUST be negative: choose (c).

Stuart Kovinsky wrote:If x^2<1, then -1 < x < 1 (in other words, x is a negative fraction, 0 or a positive fraction).

Great explanation Stuart. I always really find your explanations are the most easiest ones to understand. But one thing I don't understand from this problem is how you say -1 < x < 1 = x ^ 2? Could you explain it pls?

Thanks once again.

to proove: x<0

stem1 : x^3 (1-x^2) < 0
if the above statement has to be true, either
x^3 or (1-x^2) has to be negative.
we dont have a clue. INSUFF

stem2 : x^2-1<0
cannot say if x<0, no clues INSUFF

Now combining both 1 and 2:

you can re-arrange the equation 2 as
1 -x^2 > 0 => multiplying by - changes the inequality

Apply to the hypothesis:
>>
if the above statement has to be true, either
x^3 or (1-x^2) has to be negative.
>>

since 1-x^2 is greater than 0, it cannot be negative.
So x^3 must be negative, hence X has to be negative
SUFF

rabab wrote:

Stuart Kovinsky wrote:If x^2<1, then -1 < x < 1 (in other words, x is a negative fraction, 0 or a positive fraction).

Great explanation Stuart. I always really find your explanations are the most easiest ones to understand. But one thing I don't understand from this problem is how you say -1 < x < 1 = x ^ 2? Could you explain it pls?

Thanks once again.

Hmm.. I'm not sure where I said that!

SimplyJat you need to get your language right