Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Tips and Tricks: Inequalities [#permalink]
14 Apr 2013, 08:20

54

This post received KUDOS

37

This post was BOOKMARKED

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:

Attachment:

eq.jpg [ 27.79 KiB | Viewed 24634 times ]

This works with \(\geq{}\) and \(\leq{}\) as well, you just add the = sign, that's it.

The METHOD I) replace the\(>,<\) sign with a more friendly \(=\) II) find \(x1\) , \(x2\) as you would do normally III) use my trick to define the interval.

If the sign of \(a\) and the operator are "the same" (>,+) or (<,-) we take the ESTERNAL values. Otherwise the INTERNAL values.

I find this method very easy to remember and to use: I'm sure it will save you time. 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).

Attachment:

a.jpg [ 51.24 KiB | Viewed 24669 times ]

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 the internal values. This same principle can be applyied to a negative \(a\)

Some examples: \(f(x)=x^2+2x-3>0\) 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 \(x<-3\) \(f(x)=x^2+2x-3<0\) same function diff operator all the steps are the same Step 3)use the "tip": \(a\) is +ve and the operator is < solution \(-3<x<1\)

Re: Tips and Tricks: Inequalities [#permalink]
15 May 2013, 02:31

13

This post received KUDOS

1

This post was BOOKMARKED

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Tips and Tricks: Inequalities [#permalink]
13 Jun 2013, 08:23

So, what you are saying is as follows:

For #1, the roots are both positive (1, 2) and thus are greater than (>) zero. The inequality itself is greater than zero, therefore we take the external values (x<1 & x>2)

For #2, the roots are both positive (1, 2) and thus are greater than (>) zero. However the inequality here is less than zero, therefore we take the values between the two toots (1<x<2)

But what happens if one root is positive and one root is negative?

Re: Tips and Tricks: Inequalities [#permalink]
13 Jun 2013, 08:27

WholeLottaLove wrote:

So, what you are saying is as follows:

For #1, the roots are both positive (1, 2) and thus are greater than (>) zero. The inequality itself is greater than zero, therefore we take the external values (x<1 & x>2)

For #2, the roots are both positive (1, 2) and thus are greater than (>) zero. However the inequality here is less than zero, therefore we take the values between the two toots (1<x<2)

But what happens if one root is positive and one root is negative?

Thanks!

. Let me know

Not quite.

I am refering to the sign of a in +-a\(x^2+bx+c\), not to the sign of the roots. You have to look at the original equation.

If that sign and the operator are the "same"(+ with >, - with <), take the exsternal values; if the are different, take the internal values. _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Tips and Tricks: Inequalities [#permalink]
26 Jul 2013, 15:38

Zarrolou wrote:

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:

Attachment:

eq.jpg

This works with \(\geq{}\) and \(\leq{}\) as well, you just add the = sign, that's it.

The METHOD I) replace the\(>,<\) sign with a more friendly \(=\) II) find \(x1\) , \(x2\) as you would do normally III) use my trick to define the interval.

If the sign of \(a\) and the operator are "the same" (>,+) or (<,-) we take the ESTERNAL values. Otherwise the INTERNAL values.

I find this method very easy to remember and to use: I'm sure it will save you time. 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).

Attachment:

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 the internal values. This same principle can be applyied to a negative \(a\)

Some examples: \(f(x)=x^2+2x-3>0\) 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 \(x<-3\) \(f(x)=x^2+2x-3<0\) same function diff operator all the steps are the same Step 3)use the "tip": \(a\) is +ve and the operator is < solution \(-3<x<1\)

If you like the tip and you're gonna use it, give it a KUDOS!

Hope you guys find it useful, regards

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 ? _________________

Yogi Bhajan: If you want to learn a thing, read that; if you want to know a thing, write that; if you want to master a thing, teach that. This message transmitted on 100% recycled electrons.

Re: Tips and Tricks: Inequalities [#permalink]
26 Jul 2013, 21:13

1

This post received KUDOS

Expert's post

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 ?

For any quadratic equation \(ax^2\) + bx + c = 0, we almost always get two roots.(say x1 and x2) The internal value means the values of x that falls between x1 and x2. And the external value mean the values of x that are greater than x2 ox less than x1 (Assuming that x1 < x2).

For example, the quadratic equation \(x^2\) - 8x + 12 = 0 will give you the roots as 2 and 6.

we get 3 intervals of x. i) x < 2 ii) 2 < x < 6 iii) x > 6 of which values that are beyond the boundaries of roots are external values i.e. x < 2 and x > 6 and that are within the boundaries of roots are internal values 2 < x < 6. Now to determine which interval to select, use the method presented by Zarrolou.

Re: Tips and Tricks: Inequalities [#permalink]
27 Jul 2013, 00:02

2

This post received KUDOS

1

This post was BOOKMARKED

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 ?

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

Attachments

Imm.JPG [ 53.54 KiB | Viewed 21987 times ]

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Tips and Tricks: Inequalities [#permalink]
27 Jul 2013, 13:48

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 ?

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

Thank you for the explanation. This is more clear to me now. The above is a quadratic equation with one variable. This may seem a very basic question as I have not plotted the quadratic equation myself yet. But can we plot an equation with one variable on a two variable axis ? _________________

Yogi Bhajan: If you want to learn a thing, read that; if you want to know a thing, write that; if you want to master a thing, teach that. This message transmitted on 100% recycled electrons.

Re: Tips and Tricks: Inequalities [#permalink]
27 Jul 2013, 22:17

3

This post received KUDOS

hb wrote:

Thank you for the explanation. This is more clear to me now. The above is a quadratic equation with one variable. This may seem a very basic question as I have not plotted the quadratic equation myself yet. But can we plot an equation with one variable on a two variable axis ?

You can plot something like

\(y=x^2+5x-15\)=> it's a parabola.

You then can read the graph in this way: 1)where it intersects the x-axis the equation has its root(s). 2)if you are asked \(y=x^2+5x-15>0\), you'll pick the "parts"/range of values for x where the graph is positive or over the y-axis. 3)if you are asked \(y=x^2+5x-15<0\), you'll pick the part where the graph is below the y-axis. 4)if you are asked \(y=x^2+5x-15=0\), you'll pick the single point(s) where the graph intersect the y axis. _________________

It is beyond a doubt that all our knowledge that begins with experience.

(1) x^2 - 4x + 3 < 0 The roots are 3 and 1. Because the sign of \(x^2\) is +, and we have \(<\) ==> INTERNAL values. \(1<x<3\), but we are told that x is an integer, so \(x\) is \(2\). Sufficient

(2) x^2 + 4x +3 > 0 The roots are -1 and -3. Because the sign of \(x^2\) is +, and we have \(>\) ==> EXTERNAL values. \(x>-1\) or \(x<-3\), this is not sufficient.

If x is an integer, is |x| > 1? Or: is x>1 or x<-1 ?

(1) (1 - 2x)(1 + x) < 0 The roots are \(-1\) and \(\frac{1}{2}\). Because if you expand the expression the sign of \(x^2\) is negative, and we have \(<\) ==> External values, so \(x<-1\) or \(x>\frac{1}{2}\). This is not sufficient to answer the question.

(2) (1 - x)(1 + 2x) < 0 The roots are \(1\) and \(-\frac{1}{2}\). Because if you expand the expression the sign of \(x^2\) is negative, and we have \(<\) ==> External values,\(x<-\frac{1}{2}\) or \(x>1\). This is not sufficient to answer the question as well.

Combining the two intervals we obtain the common part, \(x<-1\) or \(x>1\) hence the answer is C (as this directly answers the main question). _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Tips and Tricks: Inequalities [#permalink]
19 Sep 2013, 03:44

This and many more tricks which I discovered here on GMAT Club off late are turning out to be a life saver. GMAT in 10 days and am wondering why wasn't I here earlier

Re: Tips and Tricks: Inequalities [#permalink]
19 Sep 2013, 03:58

Expert's post

3

This post was BOOKMARKED

violetsplash wrote:

This and many more tricks which I discovered here on GMAT Club off late are turning out to be a life saver. GMAT in 10 days and am wondering why wasn't I here earlier

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

I am not panicking. Nope, Not at all. But I am beginning to wonder what I was thinking when I decided to work full-time and plan my cross-continent relocation...

Over the last week my Facebook wall has been flooded with most positive, almost euphoric emotions: “End of a fantastic school year”, “What a life-changing year it’s been”, “My...