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# Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x

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Math Expert
Joined: 02 Sep 2009
Posts: 55230
Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

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05 Sep 2017, 01:30
00:00

Difficulty:

45% (medium)

Question Stats:

73% (01:32) correct 28% (01:46) wrong based on 60 sessions

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Is x a positive number?

(1) (x – 2)^2 > 2
(2) 2^x > 3^x

_________________
Current Student
Joined: 02 Jul 2017
Posts: 294
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

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05 Sep 2017, 01:45
x a positive number?

(1) (x – 2)^2 > 2
Assume x=1 (1-2)^2 > 2 No...
but x=0.1 positive (0.1-2)^2 = (1.9)^2 > 2... True.....
Not sufficient

(2) 2^x > 3^x
Assume x=1 => 2>3.. False
x=0.1 => 2^0.1 > 3^0.1 => False.. All positive powers will give answer "False"
so for 2^x > 3^x ... x is not positive
Sufficient

Director
Joined: 21 May 2013
Posts: 656
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

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05 Sep 2017, 02:16
Bunuel wrote:
Is x a positive number?

(1) (x – 2)^2 > 2
(2) 2^x > 3^x

Bunuel,

I have a doubt here.

For statement 1,
I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 55230
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

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05 Sep 2017, 02:28
KS15 wrote:
Bunuel wrote:
Is x a positive number?

(1) (x – 2)^2 > 2
(2) 2^x > 3^x

Bunuel,

I have a doubt here.

For statement 1,
I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks

Well yes but it will give the same answer.

Probably the easiest way to deal with the first statement would be to tests numbers or to take the square root from both sides:

$$|x-2|>\sqrt{2}$$;

$$x-2>\sqrt{2}$$ or $$-(x-2)>\sqrt{2}$$

$$x>2+\sqrt{2}$$ or $$x<2-\sqrt{2}$$
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Director
Joined: 21 May 2013
Posts: 656
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

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05 Sep 2017, 02:34
Bunuel wrote:
KS15 wrote:
Bunuel wrote:
Is x a positive number?

(1) (x – 2)^2 > 2
(2) 2^x > 3^x

Bunuel,

I have a doubt here.

For statement 1,
I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks

Well yes but it will give the same answer.

Probably the easiest way to deal with the first statement would be to tests numbers or to take the square root from both sides:

$$|x-2|>\sqrt{2}$$;

$$x-2>\sqrt{2}$$ or $$-(x-2)>\sqrt{2}$$

$$x>2+\sqrt{2}$$ or $$x<2-\sqrt{2}$$

Can you show this using the formula a^2-b^2 i.e (a-b)(a+b) and then solving? I know it is easy to solve using numbers but not able to get using this formula

Thanks
Current Student
Joined: 02 Jul 2017
Posts: 294
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

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Updated on: 05 Sep 2017, 02:52
KS15 wrote:
Can you show this using the formula a^2-b^2 i.e (a-b)(a+b) and then solving? I know it is easy to solve using numbers but not able to get using this formula

Thanks

KS15

(x – 2)^2 > 2
=> (x – 2)^2 - 2 >0
=>$$(x – 2)^2 - \sqrt{2} >0$$
=> $$(x-2+\sqrt{2}) (x-2-\sqrt{2}) >0$$

This will term either both term are positive or both terms are negative
both positive :
$$x-2+\sqrt{2} > 0$$ and $$x-2-\sqrt{2} > 0$$
$$x>2-\sqrt{2}$$ and $$x >2+\sqrt{2}$$

OR

both negative :
$$x-2+\sqrt{2} < 0$$ and $$x-2-\sqrt{2} < 0$$
$$x<2-\sqrt{2}$$and $$x <2+\sqrt{2}$$

Combining all 4 equations we can write :

$$x >2+\sqrt{2}$$ OR $$x<2-\sqrt{2}$$

Originally posted by Nikkb on 05 Sep 2017, 02:42.
Last edited by Nikkb on 05 Sep 2017, 02:52, edited 3 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 55230
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x  [#permalink]

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05 Sep 2017, 02:45
KS15 wrote:
Bunuel wrote:
KS15 wrote:

Bunuel,

I have a doubt here.

For statement 1,
I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks

Well yes but it will give the same answer.

Probably the easiest way to deal with the first statement would be to tests numbers or to take the square root from both sides:

$$|x-2|>\sqrt{2}$$;

$$x-2>\sqrt{2}$$ or $$-(x-2)>\sqrt{2}$$

$$x>2+\sqrt{2}$$ or $$x<2-\sqrt{2}$$

Can you show this using the formula a^2-b^2 i.e (a-b)(a+b) and then solving? I know it is easy to solve using numbers but not able to get using this formula

Thanks

$$(x – 2)^2 > 2$$

$$(x – 2)^2 -2> 0$$

$$(x – 2-\sqrt{2})(x – 2+\sqrt{2})> 0$$.

For the product to be positive both multiples must have the same sign.

When both are positive:
$$x – 2-\sqrt{2}> 0$$ and $$x – 2+\sqrt{2}> 0$$.
$$x > 2+\sqrt{2}$$ and $$x > 2-\sqrt{2}$$.

For both above to be true simultaneously, $$x > 2+\sqrt{2}$$ should hold (so x must be more than the larger number).

When both are negative:
$$x – 2-\sqrt{2}< 0$$ and $$x – 2+\sqrt{2}< 0$$.
$$x < 2+\sqrt{2}$$ and $$x < 2-\sqrt{2}$$.

For both above to be true simultaneously, $$x < 2-\sqrt{2}$$ should hold (so x must be less than the smaller number).

So, $$x>2+\sqrt{2}$$ or $$x<2-\sqrt{2}$$.

Hope it's clear.
_________________
Re: Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x   [#permalink] 05 Sep 2017, 02:45
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