Hi guys, I created this topic to share with you a quick and easy way to solve inequalities (second-degree). Straightaway here's my tip: eq.jpg This works with \geq{} and \leq{} as well, you just add the = sign, that's it. The METHOD I) replace the >, ,+) or ( This is it, but for those of you interested in what goes on behind the scene I have a mathematical explanation. ax^2+bx+c is a parabola. I took two parabolas one with a positive a , one with a negative a . a in the formulas defines 2 characteristics of the parabola: 1) its "slope" ( greater a greater the slope) 2) where the parabola looks at ( a +ve the parabola looks up, a -ve the parabola looks down). a.jpg With this little theory, and looking to the graphs we can conclude that: if a is +ve and we want the values >0 we have to take the esternal values , if we want the values 0 Step 1) replace > with = : x^2+2x-3 = 0 Step 2) find x1 and x2: x1=1, x2=-3 Step 3) use the "tip": a is +ve and the operator is > solution x>1 and x tips-and-tricks-inequalities-150873.html#p1225182 If you like the tip and you're gonna use it, give it a KUDOS! Hope you guys find it useful, regards

If the sign of the inequality is negative, then you have to go with negative interval. For more info refer these articles Part 1 , Part 2 In that case, you can move negative x^2 to other side of the inequality so as to make it positive and then can solve as per the earlier method. Hope that helps!

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Tips and Tricks: Inequalities

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TirthankarP

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22 Sep 2013, 04:19

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Narenn

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Bunuel

range-for-variable-x-in-a-given-inequality-109468.html

Zarrolou,

In this particular question we can see that there 4 possible ranges : \(x<-\frac{1}{3}, -\frac{1}{3}<x<0, 0<x<\frac{2}{5} and x>\frac{2}{5}\)

So how is your method applicable here?

\(15x - \frac{2}{x} > 1\) -------> \(\frac{15x^2 - 2}{x} - 1 > 0\) -----------> \(\frac{15x^2 - x - 2}{x} > 0\) -----------> \(\frac{(3x+1)(5x-2)}{x} > 0\)

The Critical points for numerator -1/3, 2/5
Critical point for denominator 0

Total critical points ------------- -1/3 --------------- 0 --------------- 2/5 ---------------

Since the sign of original inequality is positive, the expression will be positive in the rightmost region and in other regions it will be alternatively negative and positive.

That means ---------------- -1/3 +++++++++ 0 ---------------- 2/5 ++++++++++++

Hence Solution of the inequality is -1/3 < x 0 , x > 2/5

Hope that helps!

If the sentence marked in red is a rule then what will happen if the inequality is negative?
And does this rule has any impact if the sign of the coefficient of \(x^2\) changes (+15 in this case)?

Narenn

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TirthankarP

If the sentence marked in red is a rule then what will happen if the inequality is negative?

If the sign of the inequality is negative, then you have to go with negative interval. For more info refer these articles Part 1, Part 2

TirthankarP

And does this rule has any impact if the sign of the coefficient of \(x^2\) changes (+15 in this case)?

In that case, you can move negative \(x^2\) to other side of the inequality so as to make it positive and then can solve as per the earlier method.

Hope that helps!

anishasjkaul

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03 May 2014, 06:25

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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?

SirBolly

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Do inequalities on the GMAT only refer to integers?

Bunuel

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19 May 2014, 02:27

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SirBolly

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.

mittalg

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19 May 2014, 20:25

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anishasjkaul

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!!!

nitesh06

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Really helpful post.. Kudos +1...

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Zarrolou

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.
Ranaazad

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20 May 2018, 05:12

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connexion

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?

abhimahna

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gmatprep001

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?

Metwing

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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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