Tips and Tricks: Inequalities

Question

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 >,< 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. 8-)

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 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 -3tips-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

Zarrolou · Accepted Answer

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 10 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}2 Hope it's clear now. Let me know

Bunuel · Answer

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

Zarrolou · Answer

If you have any question or doubt, please ask or send me a PM!

Narenn · Answer

Thanks Dear for putting all this here.

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

Regards,

Narenn

connexion · Answer

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

Thanks.

doe007 · Answer

Zarrolou, this is great indeed. Only one small matrix to remember once the fundamental is clear! Your information deserves multiple kudos.

WholeLottaLove · Answer

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

Zarrolou · Answer

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.

hb · Answer

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 ?

Narenn · Answer

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.

Zarrolou · Answer

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

hb · Answer

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 ?

Zarrolou · Answer

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.

GMAT40 · Answer

Hi Zarrolou,

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

Zarrolou · Answer

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 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).

GMAT40 · Answer

Hi Zarrolou,

Many Thanks

Will post some questions if i get stuck...

violetsplash · Answer

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

TirthankarP · Answer

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?

Narenn · Answer

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!

Tips and Tricks: Inequalities

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Tips and Tricks: Inequalities

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