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World Problem

by phoenix9801 » Sun May 27, 2012 9:08 am
1st Question:

Can anyone tell me what is the difference when you have (x^2) vs (x)^2 ??????




2nd Question:

Also If plug in -4 or 4 into this equation it will give me the same result x^-2 < x^2 correct ????
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by aneesh.kg » Sun May 27, 2012 10:45 am
phoenix9801 wrote:1st Question:

Can anyone tell me what is the difference when you have (x^2) vs (x)^2 ??????

2nd Question:

Also If plug in -4 or 4 into this equation it will give me the same result x^-2 < x^2 correct ????
Answer 1: They just look different. Mathematically, (x^2) and (x)^2 are the same.
Answer 2: Whether you plug -4 or 4 in (x)^-2 < (x)^2, you will get the same result. The inequality holds true for both those values.
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by phoenix9801 » Sun May 27, 2012 11:30 am
Thank You, But I have one more question in mind, So how can I solve this problem using -4 and 4 ?????????


Which of the following statements must be true?

Indicate all such statements.

a) (square root(x))^3 = x^-1
b) (x^2) = (x)^2
c) x^-2 < x^2


aneesh.kg wrote:
phoenix9801 wrote:1st Question:

Can anyone tell me what is the difference when you have (x^2) vs (x)^2 ??????

2nd Question:

Also If plug in -4 or 4 into this equation it will give me the same result x^-2 < x^2 correct ????
Answer 1: They just look different. Mathematically, (x^2) and (x)^2 are the same.
Answer 2: Whether you plug -4 or 4 in (x)^-2 < (x)^2, you will get the same result. The inequality holds true for both those values.
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by vickythakre » Sun May 27, 2012 11:46 am
aneesh.kg wrote:
phoenix9801 wrote:1st Question:

Can anyone tell me what is the difference when you have (x^2) vs (x)^2 ??????

2nd Question:

Also If plug in -4 or 4 into this equation it will give me the same result x^-2 < x^2 correct ????
Answer 1: They just look different. Mathematically, (x^2) and (x)^2 are the same.
Answer 2: Whether you plug -4 or 4 in (x)^-2 < (x)^2, you will get the same result. The inequality holds true for both those values.
Aneesh,

for second question.
x^-2 < x2

LHS : 1/x^2
Which will become 1/ (-4) ^ 2
i.e. 1 / -1 * (4^2)
i.e. 1/-16
RHS : (-4)^2
i.e. -1* (4)^2
i.e. -16

Now LHS = -0.0625
& RHS = -16
If i am not wrong then LHS > RHS.

Please correct me if i am wrong.
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by aneesh.kg » Sun May 27, 2012 11:58 am
vickythakre wrote:
Aneesh,

for second question.
x^-2 < x2

LHS : 1/x^2
Which will become 1/ (-4) ^ 2
i.e. 1 / -1 * (4^2)
i.e. 1/-16
RHS : (-4)^2
i.e. -1* (4)^2
i.e. -16

Now LHS = -0.0625
& RHS = -16
If i am not wrong then LHS > RHS.

Please correct me if i am wrong.
Hi,

You've made a mistake there.

1/(-4)^2 = 1/(-4)*(-4) = 1/16
and
(-4)^2 = (-4)*(-4) = 16
are correct.
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by aneesh.kg » Sun May 27, 2012 12:07 pm
phoenix9801 wrote: Which of the following statements must be true?

Indicate all such statements.

a) (square root(x))^3 = x^-1
b) (x^2) = (x)^2
c) x^-2 < x^2
a)
(x)^3/2 = 1/x
(x)^5/2 = 1
This works out only for x = 1 and not for any other value of x.

b)
LHS and RHS are the same thing.

c)
(1/x^2) < x^2
If you plug-in values, you will realise that the inequality does not hold true for values of x between -1 and 1 such as x = 1/2 and x = -1/2.

May I please know the source of these problems? They don't look like well-made GMAT problems, especially the second statement.
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by phoenix9801 » Sun May 27, 2012 2:03 pm
Hi, Sure It a book for Standardized Test like Gmat and etc.. It also teach how Gmat can play with inequalities, word, problems, etc.... Main idea is to have a strong foundation. Thats is all. Let me know if you have any other questions.

Here is the actual Problem:

Which of the following statements must be true?

Indicate all such statements.

a) (square root(x))^3 = x^-1
b) (x^2) = (x)^2
c) x^-2 = x^2

I know for the Choice A is not correct because you can't have a negative inside a square root and the result will never be the same.

I know for The Choice B will always be True because of the cube power. Hence, It should be the only answer.


But When it come to choice C and plug in -4 and try with positive 4. it always comes true. Which become confusing. Maybe I am incorrect. If so can you show me how is it different when plugging in -4 and 4 step by step instructions please. If the statement is incorrect or correct. The way you explained it still confusing to me.

I really appreciate your help and time. If not too much to ask please explain it clearly to me. Thanks




aneesh.kg wrote:
phoenix9801 wrote: Which of the following statements must be true?

Indicate all such statements.

a) (square root(x))^3 = x^-1
b) (x^2) = (x)^2
c) x^-2 < x^2
a)
(x)^3/2 = 1/x
(x)^5/2 = 1
This works out only for x = 1 and not for any other value of x.

b)
LHS and RHS are the same thing.

c)
(1/x^2) < x^2
If you plug-in values, you will realise that the inequality does not hold true for values of x between -1 and 1 such as x = 1/2 and x = -1/2.

May I please know the source of these problems? They don't look like well-made GMAT problems, especially the second statement.
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by neelgandham » Sun May 27, 2012 5:23 pm
a) (square root(x))^3 = x^-1
b) (x^2) = (x)^2
c) x^-2 = x^2

I will try to help you with this!
a) (square root(x))^3 = x^-1
(√x)^3 = x^-1
We know that √x = x^(1/2)
So, (x^(1/2))^3 = x^-1.
So, x^(3/2) = x^-1 [Because(x^a)^b) = x^(ab)]
If x = 0 then 0^(3/2) = 0^-1 => 0 = 0. So, (√x)^3 = x^-1.
If x = 4 then 4^(3/2) = 4^-1 => 8 = 1/4. No! So, (√x)^3 is not always equal to x^-1.
Statement A doesn't satisfy the 'MUST BE TRUE' condition.
b) (x^2) = (x)^2
(x^2) = (x)^2
(x)^2 = (x)^2. In simpler terms, both the terms mean the same.
If x = 0 then (0^2) = (0)^2 => 0 = 0. So, (x^2) = (x)^2.
If x = 4 then (4^2) = (4)^2 => 16 = 16.(x^2) = (x)^2.
Statement B satisfies the 'MUST BE TRUE' condition.
c) x^-2 = x^2
x^-2 = x^2
1/(x^2) = x^2[Because, x^-a = (1/x)^a]
1 = (x^2) * (x^2)
1 = (x^4) [Because, (x^a) * (x^b) = x^(a+b)]
If x = 1 then 1 = (1^4) => 1 = 1. So, x^-2 = x^2.
If x = 4 then 1 = (4^4) => 1 = 256. So, x^-2 is not equal to x^2.
If x = -4 then 1 = (-4^4) => 1 = 256. So, x^-2 is not equal to x^2.
Statement C doesn't satisfy the 'MUST BE TRUE' condition.

I hope that answers your question.

x^-2 = x^2

If x = 4,
then x^-2 = 4^-2 = (1/4)^2 = 1/16 and
then x^2 = 4^2 = 16
So the value of x^-2 is definitely not equal to x^2.

If x = -4,
then x^-2 = (-4)^-2 = (1/-4)^2 = 1/16 and
then x^2 = (-4)^2 = 16
So the value of x^-2 is definitely not equal to x^2
Anil Gandham
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by aneesh.kg » Sun May 27, 2012 8:38 pm
You're right. As I discussed in the posts above, the (c) statement works for both 4 and -4. However, the question is asking us if this statement 'must be true'. So it is not enough for the statement to work out only for a few values. Only if it works out for all the values of x can we say that it 'must be true'.

(x)^-2 < (x)^2 does not hold true for x = 1/2 and x = - 1/2:
(1/2)^-2 = 1/(1/2)^2 =(2)^2 = 4 > (1/2)^2 = 1/4
and
(-1/2)^-2 = 1/(-1/2)^2 = (-2)^2 = 4 > (1/2)^2 = 1/4
And this is enough for us to rule this statement out.

Does that clarify your doubt?
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