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Simple inequality Question..need help

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Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 09:26
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Im not sure why I dont see this...I saw this in an answer explanation

x^2 > x breaks down to x < -1 or x > 1

i can break this apart to be x(x - 1) > 0 So x > 0 or x > 1 . Now theoretically i understand that any negative number squared, or to an even power, is positive but i dont see how x < -1 is derived from that inequality

help is appreciated. thanks
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 11:19
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sac8513 wrote:
Im not sure why I dont see this...I saw this in an answer explanation

x^2 > x breaks down to x < -1 or x > 1

i can break this apart to be x(x - 1) > 0 So x > 0 or x > 1 . Now theoretically i understand that any negative number squared, or to an even power, is positive but i dont see how x < -1 is derived from that inequality

help is appreciated. thanks


Hi,

Whenever we get a quadratic equation (or any polynomial for that matter) in an inequality, we calculate the roots merely to find out at which points on the x-axis (considering a polynomial in 'x') the equation will be equal to 0. These points at which at which the equation is 0, help us to identify the different intervals over which we have to check the consistency of the equation. Confusing? Let me explain it with the help of the example you have provided.

Considering the quadratic equation : x^2 - x > 0

First we will have to find the roots of this equation which you have rightly calculated to be 0 and 1.

Now on the x-axis, plot these two values of 'x' for which the equation becomes 0. This would divide the x-axis into three intervals. Namely :
1) x < 0
2) 0 < x < 1
3) x > 1

Now we have to check the consistency of each of these intervals by picking any value contained in the interval and seeing whether it satisfies our original inequality : x^2 - x > 0

Interval 1 : x < 0
Lets pick x = - 1
(-1)^2 - (-1) > 0 ---> 2 > 0
Since it satisfies our inequality, x must be true for all x < 0

Interval 2 : 0 < x < 1
Lets pick x = 0.5
(0.5)^2 - 0.5 > 0 ---> -0.25 > 0
Since this is false, x cannot be true for all 0 < x < 1

Interval 3 : x > 1
Lets pick 2
(2)^2 - 2 > 0 ---> 2 > 0
Since it satisfies our inequality, x must be true for all x > 1

Thus the range of x such that x satisfies the inequality x^2 - x > 0 is :
1) x < 0
2) x > 1


This is the method for solving any inequality dealing with polynomials.

Note : The equation is satisfied for all negative values of x and not just those less than -1 as you have mentioned in your post.

A quicker method that I use to find out the range of 'x' in quadratic equations is thinking of the equation to be a parabola of the form : y = ax^2 + bx + c

Now there are a few things to note here :
1) When 'a' is positive, the parabola will open upwards.
2) When 'a' is negative, the parabola will open downwards (although you can multiply the entire inequality by -1 and make 'a' positive, resulting in a parabola that faces upwards. The end result will be the same).
3) The intersection of the parabola with the x axis will represent the two roots of the equation.

Attachment:
Untitled.png
Untitled.png [ 11.04 KiB | Viewed 1179 times ]


Again considering the example you provided : y = x^2 - x is our parabola and our inequality becomes y > 0. Thus we have to find the regions for which the value of 'y' is positive.

Now since the coefficient of x^2 is positive, the parabola will 'open upwards'. Also, the parabola will intersect the x-axis at points 0 and 1.

Attachment:
Untitled1.png
Untitled1.png [ 8.64 KiB | Viewed 1175 times ]


Since we want to find the regions for which 'y' is positive and we know that it is an upward facing parabola, it is obvious that the regions we need to consider are x < 0 and x > 1 because in the region 0 < x < 1, y will be negative.

This is a much quicker method than sitting and checking all the intervals individually.
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Last edited by sriharimurthy on 08 Dec 2009, 11:29, edited 1 time in total.
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 11:20
sac8513 wrote:
Im not sure why I dont see this...I saw this in an answer explanation

x^2 > x breaks down to x < -1 or x > 1

i can break this apart to be x(x - 1) > 0 So x > 0 or x > 1 . Now theoretically i understand that any negative number squared, or to an even power, is positive but i dont see how x < -1 is derived from that inequality

help is appreciated. thanks


x(x-1) > 0 is true but you don't know if x is positive or negative...your equation assumes x > 0 but what if x were negative

the equation becomes

-x(-x-1)>0

-x >1
x < -1
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 11:56
Ok thanks . that helps. So i just need to always think that X could be negative whenever i see X^2 anywaher in an inequality.
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 12:52
sac8513 wrote:
Ok thanks . that helps. So i just need to always think that X could be negative whenever i see X^2 anywaher in an inequality.


That is not the correct way of thinking

In the example you have provided, the range of x is : x < 0 and x > 1 NOT x < -1 and x > 1.
The equation holds true for any negative value of x. For example, in the range you have mentioned, x = -0.5 is not included. However, it satisfies the equation.

lagomez wrote:
x(x-1) > 0 is true but you don't know if x is positive or negative...your equation assumes x > 0 but what if x were negative

the equation becomes

-x(-x-1)>0

-x >1
x < -1


As I mentioned earlier, the range is not correct.

Also, this method of solving is not correct.

You cannot consider x to be negative, and arrive at the inequality : -x(-x-1)>0

If you have the inequality : x(x-1)>0
You have to consider whether x and (x-1) are positive. Not just whether x is positive.

This can in no way give you the range of 'x'.

Please refer to my earlier post to see how these sort of questions should be approached.
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 13:20
So is ' lagomez ' incorrect when he says :

Quote:
x(x-1) > 0 is true but you don't know if x is positive or negative...your equation assumes x > 0 but what if x were negative

the equation becomes

-x(-x-1)>0

-x >1
x < -1
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 13:47
sac8513 wrote:
So is ' lagomez ' incorrect when he says :

Quote:
x(x-1) > 0 is true but you don't know if x is positive or negative...your equation assumes x > 0 but what if x were negative

the equation becomes

-x(-x-1)>0

-x >1
x < -1


Yes. Whenever you multiply or divide an inequality with a negative number the sign of the inequality changes. Even if the right hand side is 0.

For eg.
(-5)*(-4) > 0
If we divide the inequality by either -4 or -5, we have to change the sign of the inequality.

Now compare it to x(x-1) > 0
If either one of the terms is negative and we divide by that term, then we have to change the sign of the inequality. BUT, we don't know if either is positive or negative and nor can we assume. Thus we cannot divide by either.

I have outlined the correct method for solving these type of questions in my first post. Please spend some time going through it and trying to understand the concept. It is fairly simple.

If you have any further doubts I will try my best to address them.
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2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html

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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 14:08
sac8513 wrote:
So is ' lagomez ' incorrect when he says :

Quote:
x(x-1) > 0 is true but you don't know if x is positive or negative...your equation assumes x > 0 but what if x were negative

the equation becomes

-x(-x-1)>0

-x >1
x < -1


Yes, sorry I was in a hurry and realized on my way home that my post was incorrect
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 14:25
Before making myself crazy...maybe there is a typo in my MGMAT book then and this doen not have to be harder than im making it out to be.

Im doing a data sufficiency and statement 2) says: X^2 > X

In a description of the statement in the answer section it says-

Statemnt 2) tells us that X^2 > X, so X < -1 OR X > 1


This is the question

If X does not = 0 , is (X^2 + 1) /X > Y ?

1) X = Y
2) Y > 0

I recognize the answer to be C , but as mentioned above in the answer description of this problem that is waht the book stated for a reason Statement 2 was insufficient
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 15:00
sac8513 wrote:
Before making myself crazy...maybe there is a typo in my MGMAT book then and this doen not have to be harder than im making it out to be.

Im doing a data sufficiency and statement 2) says: X^2 > X

In a description of the statement in the answer section it says-

Statemnt 2) tells us that X^2 > X, so X < -1 OR X > 1


This is the question

If X does not = 0 , is (X^2 + 1) /X > Y ?

1) X = Y
2) Y > 0

I recognize the answer to be C , but as mentioned above in the answer description of this problem that is waht the book stated for a reason Statement 2 was insufficient


Best thing to do is when you see a problem like that ask yourself two things:
1. Is there any mention of the word integer
2. Is there any mention of positive or negative

If no to both then 90% chance you'll need to test positives and negatives and/or fractions
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 15:10
Thanks , Thats very good advice.

As far as the x < -1 . Would you say that is a typo Or hae any relation to that problem?
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 15:19
sac8513 wrote:
Thanks , Thats very good advice.

As far as the x < -1 . Would you say that is a typo Or hae any relation to that problem?


Yes, it relates to the problem.

Let's say x>1

3^2 - 1 /3 = 8/3
Is that greater than y, maybe depending on y, so answer could be yes or no

but what if x < -1

-2^2 - 1/-2 = 3/-2 and we know y is positive so the answer would be no

so knowing x < -1 very relevant
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Re: Simple inequality Question..need help [#permalink] New post 08 Dec 2009, 15:42
I defintiely get that. I guess im saying algebraically how can that be concluded.

and if X is negative . Why cant it just be stated as X < 0
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Re: Simple inequality Question..need help [#permalink] New post 09 Dec 2009, 01:46
sac8513 wrote:
I defintiely get that. I guess im saying algebraically how can that be concluded.

and if X is negative . Why cant it just be stated as X < 0


I would say its probably a typo. However, not being sure of the exact context in which they might have stated it, I wouldn't really want to comment.

What I can do though is illustrate my approach to this question. You might find it easier and more convincing than what you've probably come across in the book.

Question Stem : Is \frac{x^2+1}{x}>y ?
The question stem can be rewritten as : Is x + \frac{1}{x} > y ?
Answers should be in the form of Yes/No.

St.(1) : x = y
Looking at the rewritten form of the question stem, it is obvious that if x and y are both positive, then the answer must be Yes.
However, if x and yare both negative, then the answer to the question stem must be No.

Since this statement gives us contradicting solutions, it cannot be sufficient.

St.(2) : y > 0
In this case, it is possible for x and yto have values such that x + \frac{1}{x} is less than, equal to or greater than y while still satisfying this statement.

Since this statement gives us contradicting solutions, it cannot be sufficient.

St.(1) and St.(2) together : x = y and y > 0
Now, if x = y and y > 0 (which also implies x>0), looking at our rewritten form of the question stem, it becomes obvious that the answer will always be Yes.

Thus St.(1) and St.(2) taken together are sufficient.

Answer : C

If this is the same way you solved this question, then there is nothing wrong with your logic and if I were you I wouldn't look too much into the MGMAT reasoning for this question.
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Re: Simple inequality Question..need help   [#permalink] 09 Dec 2009, 01:46
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