Is x/3 + 3/x > 2 ?

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Is x/3 + 3/x > 2 ?

by Abdulla » Sun Dec 13, 2009 4:22 pm
OA is C
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by cramya » Sun Dec 13, 2009 5:03 pm
Simplify

Is x/3+3/x > 2

Is (x^2+9 ) / 3x > 2

Stmt I

x<3

x= -1 NO
x= 2 YES

INSUFF

Stmt II
x>1 (i.e. x is positive)

Now again the question is Is (x^2+9 ) / 3x > 2

Multiply by 3x on both sides (doesnt affect the sign of inequality since x is positive

Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^2-6x+9>0

Or Is (x-3) (x-3) > 0 or simply Is x>3 or x < -3

If x = 4 YES
If x= 3 NO

INSUFF

TOGETHER:

x is between 1 and 3 so we get a definite NO

C

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by mehravikas » Sun Dec 13, 2009 8:06 pm
Hi Cramya,

Can you explain how did you get x < -3?

Thanks,
Vikas
cramya wrote:Simplify

Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^2-6x+9>0

Or Is (x-3) (x-3) > 0 or simply Is x>3 or x < -3

C

Regards,
CR

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by Abdulla » Sun Dec 13, 2009 8:06 pm
cramya wrote:Simplify

Is x/3+3/x > 2

Is (x^2+9 ) / 3x > 2

Stmt I

x<3

x= -1 NO
x= 2 YES

INSUFF

Stmt II
x>1 (i.e. x is positive)

Now again the question is Is (x^2+9 ) / 3x > 2

Multiply by 3x on both sides (doesnt affect the sign of inequality since x is positive

Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^2-6x+9>0

Or Is (x-3) (x-3) > 0 or simply Is x>3 or x < -3

If x = 4 YES
If x= 3 NO

INSUFF

TOGETHER:

x is between 1 and 3 so we get a definite NO

C

Regards,
CR
Nice work Cramya, but can you solve it without plugging numbers. I need to see it algebraically please.
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by [email protected] » Sun Dec 13, 2009 10:49 pm
mehravikas wrote:Hi Cramya,

Can you explain how did you get x < -3?

Thanks,
Vikas
cramya wrote:Simplify

Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^2-6x+9>0

Or Is (x-3) (x-3) > 0 or simply Is x>3 or x < -3

C

Regards,
CR
Actually, that solution is incorrect. The second inequality should be x < 3.

We have:

Is (x-3) (x-3) > 0?

Well, the product of two terms is positive when both are positive or both are negative.

So, to see when that inequality holds true, we solve for those two cases:

I) x-3 > 0
x > 3

II) x-3 < 0
x < 3

Basically, the answer will be "YES" for all values of x other than 3.

So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
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by Abdulla » Mon Dec 14, 2009 12:17 am
Stuart Kovinsky wrote:
mehravikas wrote:Hi Cramya,

Can you explain how did you get x < -3?

Thanks,
Vikas
cramya wrote:Simplify

Therefore Is (x^2+9 ) / 3x > 2 becomes Is x^2+9>6x or Is x^2-6x+9>0

Or Is (x-3) (x-3) > 0 or simply Is x>3 or x < -3

C

Regards,
CR
Actually, that solution is incorrect. The second inequality should be x < 3.

We have:

Is (x-3) (x-3) > 0?

Well, the product of two terms is positive when both are positive or both are negative.

So, to see when that inequality holds true, we solve for those two cases:

I) x-3 > 0
x > 3

II) x-3 < 0
x < 3

Basically, the answer will be "YES" for all values of x other than 3.

So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
So basically you're saying that x could equal any number but not 3 in order to make the question true?

I am confused with why statement one is not sufficient. We're given that x<3 which satisfies the equation (x-3)(x-3) > 0 ------> always yes..... so why it's insufficient.
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by [email protected] » Mon Dec 14, 2009 12:25 am
Abdulla wrote:So basically you're saying that x could equal any number but not 3 in order to make the question true?

I am confused with why statement one is not sufficient. We're given that x<3 which satisfies the equation (x-3)(x-3) > 0 ------> always yes..... so why it's insufficient.
Because, as Cramya noted, we could only perform all those manipulations once we knew that x was positive; with just statement 1, negative values of x give us a "no" answer.

We need statement (2) to affirm that x is positive.
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by [email protected] » Mon Dec 14, 2009 10:15 am
Abdulla wrote:OA is C
I'd like to throw out a systematic way to handle quadratic (& cubic, quartic, etc) inequalities.
It allows us to do all of the work before we examine the two statements:
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by Abdulla » Mon Dec 14, 2009 11:34 am
Brent Hanneson wrote:
Abdulla wrote:OA is C
I'd like to throw out a systematic way to handle quadratic (& cubic, quartic, etc) inequalities.
It allows us to do all of the work before we examine the two statements:
Image
Thanks Brent. Useful information.
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by Abdulla » Mon Dec 14, 2009 12:08 pm
Stuart Kovinsky wrote:
Abdulla wrote:So basically you're saying that x could equal any number but not 3 in order to make the question true?

I am confused with why statement one is not sufficient. We're given that x<3 which satisfies the equation (x-3)(x-3) > 0 ------> always yes..... so why it's insufficient.
Because, as Cramya noted, we could only perform all those manipulations once we knew that x was positive; with just statement 1, negative values of x give us a "no" answer.

We need statement (2) to affirm that x is positive.
Thanks Stuart.

I think plugging numbers from the beginning without any manipulation is the best and easiest approach to solve this question.

Is x/3 + 3/x > 2 ?

1) x<3
plug 2 for x : 2/3 + 3/2 > 2 ----------> yes
plug 0 for x : 0/3 + 3/0 > 2 -----------> no (undefined)

2) x>1
Plug 2 for x : 2/3 + 3/2 > 2 ------------> yes
Plug 3 for x : 3/3 + 3/3 > 2 ------------> no

Combined: Any number between 1 and 3 will give a yes answer.
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by mehravikas » Mon Dec 14, 2009 12:49 pm
Hi Stuart,

We can say x < 3 only because we do not know whether x is +ve or -ve?

Vikas
Stuart Kovinsky wrote: We have:

Is (x-3) (x-3) > 0?

Well, the product of two terms is positive when both are positive or both are negative.

So, to see when that inequality holds true, we solve for those two cases:

I) x-3 > 0
x > 3

II) x-3 < 0
x < 3

Basically, the answer will be "YES" for all values of x other than 3.

So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.

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by [email protected] » Mon Dec 14, 2009 1:48 pm
mehravikas wrote:Hi Stuart,

We can say x < 3 only because we do not know whether x is +ve or -ve?

Vikas
Stuart Kovinsky wrote: We have:

Is (x-3) (x-3) > 0?

Well, the product of two terms is positive when both are positive or both are negative.

So, to see when that inequality holds true, we solve for those two cases:

I) x-3 > 0
x > 3

II) x-3 < 0
x < 3

Basically, the answer will be "YES" for all values of x other than 3.

So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
We can say that x < 3 (is one possibility) as soon as we get to the inequality:

(x-3) (x-3) > 0

To get to that inequality in the first place, we have to know that x is positive.
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by mehravikas » Mon Dec 14, 2009 7:09 pm
May be I am missing a rule completely but if we break down the inequality (x-3)(x-3) > 0 then

1. x - 3 > 0 -> x > 3
2. x - 3 > 0 -> x > 3

It would be the same if we have an inequality (x -1) (x + 5) > 0

therefore x > 1 or x > -5, here we can never say x > 1 and x < -5

I don't know whether I have given a good example, please don't mind if its a crap example because I am quite confused after seeing x > 3 or x < 3


Stuart Kovinsky wrote:
mehravikas wrote:Hi Stuart,

We can say x < 3 only because we do not know whether x is +ve or -ve?

Vikas
Stuart Kovinsky wrote: We have:

Is (x-3) (x-3) > 0?

Well, the product of two terms is positive when both are positive or both are negative.

So, to see when that inequality holds true, we solve for those two cases:

I) x-3 > 0
x > 3

II) x-3 < 0
x < 3

Basically, the answer will be "YES" for all values of x other than 3.

So, looking at the statements in combination, since we know that 1 < x < 3, x is not 3, giving us a definite "YES" answer.
We can say that x < 3 (is one possibility) as soon as we get to the inequality:

(x-3) (x-3) > 0

To get to that inequality in the first place, we have to know that x is positive.

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by [email protected] » Mon Dec 14, 2009 8:10 pm
Reread the logic of my first response.

If we know that x*y > 0, there are two possibilities: both x and y are positive or both x and y are negative.

Similarly in this question, in order for (x-3)(x-3) to be positive, either both terms are positive or both terms are negative.

So, there are two cases:

1) Both positive, in which case (x-3) and (x-3) are both > 0, so:

x-3 > 0
x > 3

2) Both negative, in which case (x-3) and (x-3) are both < 0, so:

x - 3 < 0
x < 3

Therefore, that inequality will hold true whenever:

x>3 OR x<3

We can apply the same logic to the example you gave:

(x -1) (x + 5) > 0

means that either:

1) (x-1) and (x+5) are BOTH positive; or
2) (x-1) and (x+5) are BOTH negative.

Solving:

1) x-1 > 0 AND x + 5 > 0

x > 1 AND x > -5

when we have two inequalities pointing in the same direction, the one that's more limiting applies, so the first case resolves to:

x > 1

2) x-1 < 0 AND x + 5 < 0

x < 1 AND x < -5

again taking the more limiting case:

x < -5

So, the original inequality does in fact mean:

x < -5 OR x > 1
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by mehravikas » Tue Dec 15, 2009 12:12 pm
Thanks a lot Stuart !!