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Inequality - Concept

by robosc9 » Thu Aug 18, 2011 8:35 pm
If the equation is |x-3| = 3-x
How does this get rephrased to x <= 3?

I understand to solve the equation we take (x-3), both positive and negative. Right?

When (x-3) is positive, x-3= 3-x
2x = 6, therefore x=3

When (x-3) is negative, -x+3=3-x
0=0?

So, the question is how the equation become x <= 3. How does '=' become '<='?

Similar issue,
From x^3(1-x^2)<0, how do we can get -1<x<0?

If I plug values, it makes sense, but not sure how to derive this using algebra. Can someone please explain?
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by kevincanspain » Thu Aug 18, 2011 11:20 pm
robosc9 wrote:If the equation is |x-3| = 3-x
How does this get rephrased to x <= 3?

I understand to solve the equation we take (x-3), both positive and negative. Right?

When (x-3) is positive, x-3= 3-x
2x = 6, therefore x=3

When (x-3) is negative, -x+3=3-x
0=0?

So, the question is how the equation become x <= 3. How does '=' become '<='?

Similar issue,
From x^3(1-x^2)<0, how do we can get -1<x<0?

If I plug values, it makes sense, but not sure how to derive this using algebra. Can someone please
explain?
You are on the right track: let's start with a definition |z| = z for all z >= 0 and |z| =-z for all z < 0

You were right to consider the cases when x-3 is positive or negative, but what about zero?

If x - 3 >= 0, we get x - 3 = 3 - x i.e. x =3

If x - 3 < 0 (i.e. x < 3) , we get 3 - x = 3 - x , which is always true, no matter the value of x on this interval

Thus x < 3 or x = 3. We write the solution set as x <=3
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by Anurag@Gurome » Fri Aug 19, 2011 4:03 am
Kevin has explained the first query, so I'm not repeating the same.
Let me shed some light on your second query.
robosc9 wrote:From x^3(1-x^2)<0, how do we can get -1<x<0?
As the product of x³ and (1 - x²) is negative, the signs of x³ and (1 - x²) must be opposite. Hence, any one of the following two cases is possible,
  • 1. x³ > 0 and (1 - x²) < 0
    2. x³ < 0 and (1 - x²) > 0
Case 1:
  • # x³ > 0 implies x > 0
    # (1 - x²) < 0 implies x² > 1, i.e. either x > 1 or x < -1
    But x must be greater than 0, hence x > 1
Case 2:
  • # x³ < 0 implies x < 0
    # (1 - x²) > 0 implies x² < 1, i.e. -1 < x < 1
    But x must be less than 0, hence -1 < x < 0
Combining all of the above, we can see that if x³(1 - x²) < 0, then either x must be greater than 1 or x must lie between -1 and 0.

Hope this helps.
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by GMATGuruNY » Fri Aug 19, 2011 6:12 am
robosc9 wrote:
Similar issue,
From x^3(1-x^2)<0, how do we can get -1<x<0?

If I plug values, it makes sense, but not sure how to derive this using algebra. Can someone please explain?
Another approach: determine the critical points -- the points at which x³(1-x²) = 0.

Factoring 1-x², we get:
x³(1+x)(1-x) < 0.
The critical points are x=0, x=-1, and x=1.
These are the only values for x at which x³(1+x)(1-x) = 0.
To determine the range of x, try one value to the left and right of each critical point.

x<-1:
Plugging x=-2 into x³(1+x)(1-x) < 0, we get:
(-2)³(1+(-2))*(1-(-2)) < 0.
-8 * -1 * 3 < 0.
24<0.
Doesn't work.
x<-1 is NOT part of the range.

-1<x<0:
Plugging x=-1/2 into x³(1+x)(1-x) < 0, we get:
(-1/2)³(1+(-1/2))*(1-(-1/2)) < 0.
(-1/8)*(1/2)*(3/2) < 0.
-3/32<0.
This works.
-1<x<0 IS part of the range.

0<x<1
Plugging x=1/2 into x³(1+x)(1-x) < 0, we get:
(1/2)³(1+(1/2))*(1-(1/2)) < 0.
(1/8)*(3/2)*(1/2) < 0.
3/32<0.
Doesn't work.
0<x<1 is NOT part of the range.

x>1
Plugging x=2 into x³(1+x)(1-x) < 0, we get:
(2)³(1+2)(1-2) < 0.
8*3*(-1) < 0.
-24<0.
This works.
x>1 IS part of the range.

Thus, -1<x<0 OR x>1.
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by robosc9 » Sat Aug 20, 2011 6:25 am
kevincanspain wrote:
You were right to consider the cases when x-3 is positive or negative, but what about zero?

If x - 3 >= 0, we get x - 3 = 3 - x i.e. x =3
Thanks, it's clear now! I have always been confused about this!
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by robosc9 » Sat Aug 20, 2011 6:34 am
GMATGuruNY wrote:
Another approach: determine the critical points -- the points at which x³(1-x²) = 0.
I like your approach too, but how do we determine these critical points and how do we decide which is a better way to solve a question (critical points/algebra)?

In my other question:
|x-3| = 3-x

Are the two critical points are x>=0 and x<3. Are there any more?
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by baladon99 » Mon Aug 29, 2011 3:25 am
robosc9 wrote:
GMATGuruNY wrote:
Another approach: determine the critical points -- the points at which x³(1-x²) = 0.
I like your approach too, but how do we determine these critical points and how do we decide which is a better way to solve a question (critical points/algebra)?

In my other question:
|x-3| = 3-x

Are the two critical points are x>=0 and x<3. Are there any more?
Modulus of any number is always positive or zero.
So R.H.S of the eqn |x-3| = 3-x is +ve or 0
3-x >= 0
x-3<= 0
x<=3
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by shankar.ashwin » Sun Nov 20, 2011 6:24 am
Mitch could you please help us on this one, after we find the critical points of an inequality and arrange them in ascending order and test for values to the left and right of these points, Does the inequality signs always alternate? I saw a few posts and in most of them they did. In this case we have the following cases;

x<-1: - Does not satisfy
-1<x<0: - Satisfies
0<x<1 - Does not satisfy
x>1 - Satisfies.

Unless we are given absolute values, does this pattern always hold true? If I am not wrong critical points are nothing but roots of the equation when the function = 0.
GMATGuruNY wrote:
Another approach: determine the critical points -- the points at which x³(1-x²) = 0.

Factoring 1-x², we get:
x³(1+x)(1-x) < 0.
The critical points are x=0, x=-1, and x=1.
These are the only values for x at which x³(1+x)(1-x) = 0.
To determine the range of x, try one value to the left and right of each critical point.
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by Ian Stewart » Mon Nov 21, 2011 3:55 am
shankar.ashwin wrote:Mitch could you please help us on this one, after we find the critical points of an inequality and arrange them in ascending order and test for values to the left and right of these points, Does the inequality signs always alternate? I saw a few posts and in most of them they did.
Most of the time they will alternate, but not always. For example, if you have this inequality, which I'll borrow from a high-level official question found in gmatfocus (so it's not some entirely irrelevant example) :

(x/3) + (3/x) > 2

There are two important things to notice here if you intend to use a critical point method.

First, the only value of x which makes (x/3) + (3/x) = 2 true is x = 3. Still, the inequality is true *both* when x > 3 and when 0 < x < 3; that is, it's true both to the left and to the right of x=3. So this is an example where the solutions do not alternate around the critical point.

Second, when finding critical points, you need to consider not only values of x which make both sides of the inequality equal, but also values of x which make one side of the inequality *undefined*. So here, since x is in a denominator, we have another critical point: x = 0. You can see how important it is to identify this critical point: if x < 0, the inequality is clearly false, and if x > 0 then the inequality is true, except when x=3.


In general, I think a critical point method is a perfectly good approach to many straightforward inequality questions which involve simple exponents or quadratics (though there are other equally good approaches you could use instead), but if you intend to use it on some of the more 'exotic' inequality questions that can appear at the high level of the test, there are complications that you may need to be aware of.
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by kumadil2011 » Mon Nov 21, 2011 9:35 pm
Hi..Could you please advise, while finding critical points, is it necessary to check range between

-1<x<0 and 0<x<1 (Proper and improper fractions, both + and - ve)In all problems?

As i missed here, concluding the range only 1<x

**************************************************************************************************
-1<x<0:
Plugging x=-1/2 into x³(1+x)(1-x) < 0, we get:
(-1/2)³(1+(-1/2))*(1-(-1/2)) < 0.
(-1/8)*(1/2)*(3/2) < 0.
-3/32<0.
This works.
-1<x<0 IS part of the range.

0<x<1
Plugging x=1/2 into x³(1+x)(1-x) < 0, we get:
(1/2)³(1+(1/2))*(1-(1/2)) < 0.
(1/8)*(3/2)*(1/2) < 0.
3/32<0.
Doesn't work.
0<x<1 is NOT part of the range.

x>1
Plugging x=2 into x³(1+x)(1-x) < 0, we get:
(2)³(1+2)(1-2) < 0.
8*3*(-1) < 0.
-24<0.
This works.
x>1 IS part of the range.

Thus, -1<x<0 OR x>1.
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by nikhilgmat31 » Wed Jul 29, 2015 9:53 pm
Another way of solving this |x-3| = 3-x

|x-3| = -(x-3)

It means x-3 is negative

x-3 <=0

x <=3
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by [email protected] » Thu Jul 30, 2015 10:44 am
Hi nikhilgmat31,

You have to be very careful with these types of deductions and the work that you do. Your work PROVES that (X-3) could equal 0, but you wrote that (X-3) is NEGATIVE (which is NOT correct). On certain questions (typically DS questions), these types of little errors will lead to an incorrect answer and could cost you some serious points on Test Day.

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by nikhilgmat31 » Thu Jul 30, 2015 9:44 pm
Right Rich, I messed up
It means x-3 is negative & x-3 <=0

induced extra = symbol :(
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by [email protected] » Fri Jul 31, 2015 5:12 pm
Hi nikhilgmat31,

It's okay to make mistakes now, but you have to treat each of these little errors as an opportunity to improve. From what I've seen of your overall work in the Quant Forum here, you know a lot of the material really well and you have to potential to score at a really high level in the Quant section on Test Day. At a certain point though, one of the 'key' elements to raising your score even further will be your 'precision' and not your intelligence. That very precision will likely come down to the amount of work that you put on the pad (enough notes, labels, details, etc. to keep the little mistakes from happening.

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by nikhilgmat31 » Mon Aug 03, 2015 4:41 am
Thanks Rich for guiding me. Sometimes to proceed faster in questions, we tend to bypass certain steps & miss variables sometimes.

Can you please provide me some good reference to counting & probability. I want to read these topics in detail.
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