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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 50284 times ]

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

 The METHODI) replace the$$>,<$$ sign with a more friendly $$=$$II) find $$x1$$ , $$x2$$ as you would do normallyIII) 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 50344 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 $$-3<x<1$$

More examples: 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

Originally posted by Zarrolou on 14 Apr 2013, 09:20.
Last edited by Zarrolou on 26 May 2013, 03:08, edited 2 times in total.
Updated
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Re: Tips and Tricks: Inequalities  [#permalink]

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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 following posts might also be helpful:
new-project-review-discuss-and-get-kudos-points-153555.html
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Re: Tips and Tricks: Inequalities  [#permalink]

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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
##### General Discussion
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Re: Tips and Tricks: Inequalities  [#permalink]

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If you have any question or doubt, please ask or send me a PM!
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Re: Tips and Tricks: Inequalities  [#permalink]

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Thanks Dear for putting all this here.

Such fundamental things, once mastered, ultimately lead us to the greater accuracy.

Regards,

Narenn
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Re: Tips and Tricks: Inequalities  [#permalink]

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Hi Zarrolou ,
Thanks for your tip. Can you please elaborate it with more examples? I have difficulty in understanding it.

Thanks.
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Re: Tips and Tricks: Inequalities  [#permalink]

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Zarrolou, this is great indeed. Only one small matrix to remember once the fundamental is clear! Your information deserves multiple kudos.
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Re: Tips and Tricks: Inequalities  [#permalink]

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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!
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Re: Tips and Tricks: Inequalities  [#permalink]

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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.
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Re: Tips and Tricks: Inequalities  [#permalink]

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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 METHODI) replace the$$>,<$$ sign with a more friendly $$=$$II) find $$x1$$ , $$x2$$ as you would do normallyIII) 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 $$-3<x<1$$

More examples: 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

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 ?
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Re: Tips and Tricks: Inequalities  [#permalink]

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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.

Hope that helps.
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Re: Tips and Tricks: Inequalities  [#permalink]

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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
Attachments Imm.JPG [ 53.54 KiB | Viewed 47508 times ]

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Re: Tips and Tricks: Inequalities  [#permalink]

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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

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 ?
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Re: Tips and Tricks: Inequalities  [#permalink]

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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.
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Re: Tips and Tricks: Inequalities  [#permalink]

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Hi Zarrolou,

Can you please explain a couple of harder DS question on inequalities.
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Re: Tips and Tricks: Inequalities  [#permalink]

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GMAT40 wrote:
Hi Zarrolou,

Can you please explain a couple of harder DS question on inequalities.

Original question : if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html

If x is an integer, what is the value of x?

(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.

A

Original question: if-x-is-an-integer-is-x-98674.html

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).
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Re: Tips and Tricks: Inequalities  [#permalink]

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Hi Zarrolou,

Many Thanks

Will post some questions if i get stuck...
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Re: Tips and Tricks: Inequalities  [#permalink]

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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 Manager  Joined: 29 Apr 2013
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Re: Tips and Tricks: Inequalities  [#permalink]

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Bunuel wrote:

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?
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Re: Tips and Tricks: Inequalities  [#permalink]

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TirthankarP wrote:
Bunuel wrote:

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! _________________
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