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# Is x/3 + 3/x > 2

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Is x/3 + 3/x > 2 [#permalink]

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04 Aug 2009, 16:33
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Is x/3 + 3/x > 2

(1) x < 3
(2) x > 1

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-3-3-x-97331.html
[Reveal] Spoiler: OA
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04 Aug 2009, 18:05
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C...

so we have x/3 + 3/x > 2...assume x > 0 (as when x < 0, equation fails, thus statement 1 alone is not true), then we have
x^2/3 + 3 - 2x > 0 => x^2 - 6x + 9 > 0 => (x-3)^2 > 0 so for all positive x except for 3 does the equation stand. thus statement 2 alone doesn't stand either...we need both. 1 < x < 3 definitely works.
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04 Aug 2009, 22:17
C ...
it is...

X= 0 is an indeterminate value ...so (i) is out

for (ii)

numbers above =>3 wont work ...

Hence both i and ii taken together is suffieient.
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04 Aug 2009, 23:43
Hi
I believe the ans is A

the stem can be re-written as (x-3)^2>0

From A we know that X<3 wnich means that this is always true since 3 is excluded which makes the stem equal to 0

A is SUFF

From 2 we know X> 1 , which means that the stem can be 0 so, it is not SUFF

Regards
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05 Aug 2009, 07:24
I think better way to look at this as an eqn a+1/a ... see the behaviour of this eqn....first image is for eqn (a+1/a) and the later is the actual behaviour of (x/3+3/x)
value of this equation cannot go below 2 (except in fractional values less than 1)..thats the property...hence OA C

If you like my post ....consider it for Kudos
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06 Aug 2009, 02:25
Is x/3+3/x>2

1 x<3

2 x>1

IMO B......

I m unable to understand explanations above...

may be I m missing somethg...OA plz

X/3+3/X > 2 = (X^2 + 9)/3X ......

1. X<3 ...equation can be postive(greater than 2) or negative..insuff

2. X>1 ..suff ..chk with any value..... let say x=2 .... (2)^2 + 9/3*2 = 13/6 >2

take x=5 ... 25+9/15 = 34/15> 2....
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07 Aug 2009, 10:55
sher676 wrote:
Is x/3+3/x>2

1 x<3

2 x>1

is (x^2+9)/3x -2>0

(x^2+9-6x)/3x>0

is (x-3)^2/3x>0 all depends on the sign of x in the deniminator

from one
insuff

from 2
suff

I d say B
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07 Aug 2009, 12:28
I would say A.

The question basically comes down to is (x-3)^2 > 0 or x > 3 ?

St. 1: x<3. Suff.
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07 Aug 2009, 16:50
yezz wrote:
sher676 wrote:
Is x/3+3/x>2

1 x<3

2 x>1

is (x^2+9)/3x -2>0

(x^2+9-6x)/3x>0

is (x-3)^2/3x>0 all depends on the sign of x in the deniminator

from one
insuff

from 2
suff

I d say B

if x = 3, the equation becomes 1 + 1 > 2 ...
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07 Aug 2009, 17:15
sprtng wrote:
Jivana wrote:
I would say A.

The question basically comes down to is (x-3)^2 > 0 or x > 3 ?

St. 1: x<3. Suff.

what if x = -1?

Still less than 3, so suff.
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07 Aug 2009, 17:31
Jivana wrote:
sprtng wrote:
Jivana wrote:
I would say A.

The question basically comes down to is (x-3)^2 > 0 or x > 3 ?

St. 1: x<3. Suff.

what if x = -1?

Still less than 3, so suff.

Is x/3+3/x>2 ?

1 x<3

so lets take two cases where x < 3...
case 1:
x = 2

2/3 + 3/2 = 13/12 > 2

case 2:
x = -2
-2/3 - 3/2 = -13/12 < 2
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07 Aug 2009, 17:50
sprtng wrote:

Is x/3+3/x>2 ?

1 x<3

so lets take two cases where x < 3...
case 1:
x = 2

2/3 + 3/2 = 13/12 > 2

case 2:
x = -2
-2/3 - 3/2 = -13/12 < 2

For case 1: 13/12 will not be greater than 2.
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07 Aug 2009, 21:45
sprtng wrote:
yezz wrote:
sher676 wrote:
Is x/3+3/x>2

1 x<3

2 x>1

is (x^2+9)/3x -2>0

(x^2+9-6x)/3x>0

is (x-3)^2/3x>0 all depends on the sign of x in the deniminator

from one
insuff

from 2
suff

I d say B

if x = 3, the equation becomes 1 + 1 > 2 ...

in this case i d say C x is defined on the domain 1<x<3
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07 Aug 2009, 22:55
sprtng wrote:
Jivana wrote:
sprtng wrote:

Is x/3+3/x>2 ?

1 x<3

so lets take two cases where x < 3...
case 1:
x = 2

2/3 + 3/2 = 13/12 > 2

case 2:
x = -2
-2/3 - 3/2 = -13/12 < 2

i meant to write 13/6....and -13/6 >. <
For case 1: 13/12 will not be greater than 2.

You are right about the x=2 situation.

Now my question is, can we right: x/3+3/x>2 as x^2 +9 -6x >0 or did I miss a step?

x/3+3/x>2
=> (x^2 + 9 ) / 3x > 2
=> x^2 + 9 > 2.3x
=> x^2 +9 -6x >0
=> (x-3)^2 > 0
or, x > 3
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08 Aug 2009, 07:26
You are right about the x=2 situation.

Now my question is, can we right: x/3+3/x>2 as x^2 +9 -6x >0 or did I miss a step?

x/3+3/x>2
=> (x^2 + 9 ) / 3x > 2
=> x^2 + 9 > 2.3x
=> x^2 +9 -6x >0
=> (x-3)^2 > 0
or, x > 3[/quote]

in this case you are assuming x > 0. and then because x > 0 and (x-3)^2 > 0, x can be anything greater than 0 but NOT 3, not necessarily n>3 only.
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23 Oct 2009, 21:26
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Expert's post
Let me try to clear out this one:

Is x/3+3/x>2
(1) x<3
(2) x>1

For a moment forget about (1) and (2).

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Nominator (x-3)^2 is NEVER negative, so expression is negative only when denominator 3x is negative, or simply when x is negative. Hence when x>0 our expression (x-3)^2/3x>0 is always positive EXCEPT when x=3, because at that case nominator (x-3)^2 becomes 0, thus expression becomes 0 and we need expression to be more then 0.

So we get that (x-3)^2/3x>0 holds true when x>0 and x#3.

Let's move to the statements:
(1) x<3, tells us that x#3, but it's not enough, we need x to be positive besides that. Not sufficient.

(2) x>1, tell us that x positive, but it's not enough we need x not to be equal to 3. Not sufficient.

(1) + (2) 1<x<3 in this range x is positive and not equal to 3. Sufficient.

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23 Oct 2009, 22:27
sprtng wrote:
You are right about the x=2 situation.

Now my question is, can we right: x/3+3/x>2 as x^2 +9 -6x >0 or did I miss a step?

x/3+3/x>2
=> (x^2 + 9 ) / 3x > 2
=> x^2 + 9 > 2.3x
=> x^2 +9 -6x >0
=> (x-3)^2 > 0
or, x > 3

step marked in RED is not correct, we cannot assume x>0 !!
hence we SHOULD NOT derive the equation (x-3)^2 > 0.

Bunuel can you please elaborate how did you arrive at this step:
x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.
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26 Oct 2009, 00:31
Economist wrote:
Bunuel can you please elaborate how did you arrive at this step:
x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Just noticed that you addressed the question. I simply reduced to a common denominator x/3, 3/x and -2:
x/3+3/x>2
--> x/3+3/x-2>0
--> (x^2+9-6x)/3>0
--> (x-3)^2/3>0
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15 Jul 2013, 15:59
Bunuel wrote:
Let me try to clear out this one:

Is x/3+3/x>2
(1) x<3
(2) x>1

For a moment forget about (1) and (2).

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Nominator (x-3)^2 NEVER negative, so expression is negative only when denominator 3x is negative, or simply when x is negative. Hence when x>0 our expression (x-3)^2/3x>0 is always positive EXCEPT when x=3, because at that case nominator (x-3)^2 becomes 0, thus expression becomes 0 and we need expression to be more then 0.

So we get that (x-3)^2/3x>0 holds true when x>0 and x#3.

Let's move to the statements:
(1) x<3, tells us that x#3, but it's not enough, we need x to be positive besides that. Not sufficient.

(2) x>1, tell us that x positive, but it's not enough we need x not to be equal to 3. Not sufficient.

(1) + (2) 1<x<3 in this range x is positive and not equal to 3. Sufficient.

Hi Bunuel,

Why do we have to keep denominator here?

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Can't i reduce this equation to (x^2-6x+9) ---> (x-3)^2 > 0.

Then with Stm1: we can prove the Stem.....
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15 Jul 2013, 22:19
Mountain14 wrote:
Bunuel wrote:
Let me try to clear out this one:

Is x/3+3/x>2
(1) x<3
(2) x>1

For a moment forget about (1) and (2).

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Nominator (x-3)^2 NEVER negative, so expression is negative only when denominator 3x is negative, or simply when x is negative. Hence when x>0 our expression (x-3)^2/3x>0 is always positive EXCEPT when x=3, because at that case nominator (x-3)^2 becomes 0, thus expression becomes 0 and we need expression to be more then 0.

So we get that (x-3)^2/3x>0 holds true when x>0 and x#3.

Let's move to the statements:
(1) x<3, tells us that x#3, but it's not enough, we need x to be positive besides that. Not sufficient.

(2) x>1, tell us that x positive, but it's not enough we need x not to be equal to 3. Not sufficient.

(1) + (2) 1<x<3 in this range x is positive and not equal to 3. Sufficient.

Hi Bunuel,

Why do we have to keep denominator here?

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Can't i reduce this equation to (x^2-6x+9) ---> (x-3)^2 > 0.

Then with Stm1: we can prove the Stem.....

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign./b]

So you CANNOT multiply $$\frac{(x-3)^2}{x}>0$$ by $$3x$$ since you don't know the sign of $$x$$: if x>0, then you'll have $$(x-3)^2>0$$ but if x<0, then you'll have $$(x-3)^2<0$$ (flip the sign when multiplying by negative value).

Hope it's clear.

[b]OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-3-3-x-97331.html

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Re: x/3   [#permalink] 15 Jul 2013, 22:19
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