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

48

This post received KUDOS

23

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 18035 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 18055 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

12

This post received KUDOS

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 +-ax^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

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 15432 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

2

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