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Re: Tips and Tricks: Inequalities [#permalink]
Do inequalities on the GMAT only refer to integers?
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Re: Tips and Tricks: Inequalities [#permalink]
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SirBolly wrote:
Do inequalities on the GMAT only refer to integers?


No. You cannot assume that a variable represents an integer unless you are explicitly told about it or unless it's obvious (for example when a variable represents the number of people, books, etc).

I suggest you to go through the following post ALL YOU NEED FOR QUANT.

Theory on Inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html


Hope this helps.
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Re: Tips and Tricks: Inequalities [#permalink]
anishasjkaul wrote:
Hi, Bunuel

The trick for inequalities written in the first post of this link, is it always true??... Will it work for all inequalities? Even third degree equation of X ( for example) with an inequality sign?



Yes, it is always true. But a more easy approach is as follows:

Lets have an inequality of degree n of the form

(x-1)(x-2)(-3-x)(x-4)<=0,

Now here, we see that coefficient of highest power of x is negative. My approach is as follows:

Step 1: Make coefficient of highest power positive by changing the sign of inequality

(x-1)(x-2)(x+3)(x-4)>=0

Step 2: Identify critical point.

Four critical points are there -3,1,2,4

Since you have already made coefficient of highest power of x positive in first step, the inequality if always be positive beyond the right most critical point i.e. beyond 4 it is always positive

Step3: Plot the regions.

Now the +ve and -ve regions will alternate


>4 it is positive
2<x<4 it is negative
1<x<2 it is positive
-3<x<1 it is negative
x<-3 it is positive.

In this way you can solve the inequality.

In case we have an even power of a term as shown below:

(x-1)(x-2)^2(x-3)(x-4)>=0

The regions at the critical point 2 will not alternate. We will again start from right most critical point:

x>4 is positive
3<x<4 is negative
2<x<3 is positive
1<x<2 is positive

x<1 is negative

In case of an odd power the signs alternate normally.

Hope it makes sense!!!

Kudos if you like!!!
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Re: Tips and Tricks: Inequalities [#permalink]
Really helpful post.. Kudos +1...
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Re: Tips and Tricks: Inequalities [#permalink]
Zarrolou wrote:
hb wrote:
Can you elaborate more on the external and internal values ? That should clear the concept completely for me. Is it possible for you to draw a shaded area on the graph that explains what you mean by the external and internal values as defined by x1 and x2 ?


Here are some more examples tips-and-tricks-inequalities-150873.html#p1225182

Say that you have \(x_1=3\) and \(x_2=5\)

External values:
------------------(3)-------------(5)------------------
Internal values:
------------(3)------------------(5)------------------

For x_1=-10 and x_2=-1
External values:
------------------(-10)-------------(-1)------------------
Internal values:
------------(-10)------------------(-1)------------------

External values=" values greater then the greatest root, and smaller than the smallest root".
Internal values="values in between the two roots".

In the image below there are two graphical examples. The first one represents the solution for
\(x^2-8x+15>0\)
The second one for
\(-x^2+8x-15>0\)

Hope everything is clear, let me know.
Thanks


Really amazing post. Although this might sound a basic question to ask, l'm not quite sure when we have to take internal values. ls it when x is negative? Could someone please clarify why we need to consider internal values on second graph? :| Would be a great help.

Regards.
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Tips and Tricks: Inequalities [#permalink]
Zarrolou wrote:
connexion wrote:
Hi Zarrolou ,
Thanks for your tip. Can you please elaborate it with more examples? I have difficulty in understanding it.

Thanks.


Sure!
Here are all possible cases:
1)\(x^2-3x+2>0\)
2)\(x^2-3x+2<0\)

The roots of \(x^2-3x+2=0\) are x=1 and x=2
In 1 the sign of x^2 is + and the operator is > => they are "the same" so we take the external values
\(x<1\) and \(x>2\)
In 2 the signs are not the same => we take the internal values
\(1<x<2\)

in these the sign of x^2 is -
3)\(-2x^2+3x+2>0\)
4)\(-2x^2+3x+2<0\)

The roots of \(-2x^2+3x+2 =0\) are \(x=2\) and \(x=-\frac{1}{2}\).
In 3 the sign of x^2 is - and the operator is > => the are not the same so we take the internal values
\(-\frac{1}{2}<x<2\)
In 4 they are the same (-,<) external values
\(x<-\frac{1}{2}\) and \(x>2\)


Hope it's clear now. Let me know


I know it's a quite old post, however let me ask one question:
Are the solution for \(-2x^2+3x+2 =0\) not \(x=-2\), i.e. negative 2, and \(x=\frac{1}{2}\), i.e. positive 1/2, hence the solution range for \(-2x^2+3x+2>0\) is -2 < x < 1/2?
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Tips and Tricks: Inequalities [#permalink]
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gmatprep001 wrote:
I know it's a quite old post, however let me ask one question:
Are the solution for \(-2x^2+3x+2 =0\) not \(x=-2\), i.e. negative 2, and \(x=\frac{1}{2}\), i.e. positive 1/2, hence the solution range for \(-2x^2+3x+2>0\) is -2 < x < 1/2?


Hey gmatprep001 ,

Let me solve the equation for you:

\(-2x^2+3x+2 =0\) can be written as \(2x^2-3x-2 =0\).

Now, \(2x^2-3x-2 =0\)

==> \(2x^2 - 4x + x - 2 =0\)

==> \(2x(x - 2) + 1 (x - 2) =0\)

==> either \(2x + 1 = 0\) or \((x - 2) =0\)

==> \(x = -1/2\) or x = 2.

Does that make sense?
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Re: Tips and Tricks: Inequalities [#permalink]
Abhimahna & gmatprep001

Even if you do not rewrite the equation, in original form also, you will get the same answer, How gmatprep001 arrived at x=-2?
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Re: Tips and Tricks: Inequalities [#permalink]
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