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Tips and Tricks: Inequalities [#permalink]
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14 Apr 2013, 09:20
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Hi guys, I created this topic to share with you a quick and easy way to solve inequalities(seconddegree). Straightaway here's my tip:
Attachment:
eq.jpg [ 27.79 KiB  Viewed 39777 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 39849 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:\(f(x)=x^2+2x3>0\) Step 1) replace > with = : \(x^2+2x3 = 0\) 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 \(x<3\) \(f(x)=x^2+2x3<0\) 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: tipsandtricksinequalities150873.html#p1225182If you like the tip and you're gonna use it, give it a KUDOS!Hope you guys find it useful, regards
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Last edited by Zarrolou on 26 May 2013, 03:08, edited 2 times in total.
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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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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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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)\(x^23x+2>0\) 2)\(x^23x+2<0\) The roots of \(x^23x+2=0\) 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
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Re: Tips and Tricks: Inequalities [#permalink]
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17 May 2013, 04:30
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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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13 Jun 2013, 09:23
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?
Thanks!



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Re: Tips and Tricks: Inequalities [#permalink]
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13 Jun 2013, 09:27
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.
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Re: Tips and Tricks: Inequalities [#permalink]
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26 Jul 2013, 16:38
Zarrolou wrote: Hi guys, I created this topic to share with you a quick and easy way to solve inequalities(seconddegree). Straightaway here's my tip: 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). 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:\(f(x)=x^2+2x3>0\) Step 1) replace > with = : \(x^2+2x3 = 0\) 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 \(x<3\) \(f(x)=x^2+2x3<0\) 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: tipsandtricksinequalities150873.html#p1225182If 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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26 Jul 2013, 22:13
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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 tipsandtricksinequalities150873.html#p1225182Say 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 \(x^28x+15>0\) The second one for \(x^2+8x15>0\) Hope everything is clear, let me know. Thanks
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Re: Tips and Tricks: Inequalities [#permalink]
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27 Jul 2013, 14:48
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 ? Here are some more examples tipsandtricksinequalities150873.html#p1225182Say 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 \(x^28x+15>0\) The second one for \(x^2+8x15>0\) 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 \(y=x^2+5x15\)=> it's a parabola. You then can read the graph in this way: 1)where it intersects the xaxis the equation has its root(s). 2)if you are asked \(y=x^2+5x15>0\), you'll pick the "parts"/range of values for x where the graph is positive or over the yaxis. 3)if you are asked \(y=x^2+5x15<0\), you'll pick the part where the graph is below the yaxis. 4)if you are asked \(y=x^2+5x15=0\), 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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31 Jul 2013, 10:09
Hi Zarrolou,
Can you please explain a couple of harder DS question on inequalities.



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GMAT40 wrote: Hi Zarrolou,
Can you please explain a couple of harder DS question on inequalities. Original question : ifxisanintegerwhatisthevalueofx1x24x94661.htmlIf x is an integer, what is the value of x?(1) x^2  4x + 3 < 0The 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. AOriginal question: ifxisanintegerisx98674.htmlIf x is an integer, is x > 1?Or: is x>1 or x<1 ? (1) (1  2x)(1 + x) < 0The 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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01 Aug 2013, 03:39
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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19 Sep 2013, 04:44
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



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Re: Tips and Tricks: Inequalities [#permalink]
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22 Sep 2013, 01:51
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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22 Sep 2013, 03:32
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? \(15x  \frac{2}{x} > 1\) > \(\frac{15x^2  2}{x}  1 > 0\) > \(\frac{15x^2  x  2}{x} > 0\) > \(\frac{(3x+1)(5x2)}{x} > 0\) 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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