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Doubt on Solution to Inequalities

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Doubt on Solution to Inequalities [#permalink] New post 24 Jul 2013, 05:58
Hi

When we take the inequality (x-3)*(x-6)>0, I understand that this eq is a parabola that passes through the X-axis at x=3 and x=6.

This also shows that the expression is positive between x>3 and x<6.

Now my question is:

We generally say the solution to the above inequality is 3<x<6.

So, if one sees (x-3)*(x-6) > 0, then is there a way to arrive at 3<x<6

or

Should we just think that as the roots are 3 and 6, and 3 being smaller, the equation has to be positive between 3 and 6.

Ex: if I say (x-a)*(x-b) > 0, I don't know whether the solution is a<x<b or b<x<a.

is there a way to arrive at this?
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Re: Doubt on Solution to Inequalities [#permalink] New post 24 Jul 2013, 07:29
Expert's post
gmatter0913 wrote:
Hi

When we take the inequality (x-3)*(x-6)>0, I understand that this eq is a parabola that passes through the X-axis at x=3 and x=6.

This also shows that the expression is positive between x>3 and x<6.

Now my question is:

We generally say the solution to the above inequality is 3<x<6.

So, if one sees (x-3)*(x-6) > 0, then is there a way to arrive at 3<x<6


or

Should we just think that as the roots are 3 and 6, and 3 being smaller, the equation has to be positive between 3 and 6.

Ex: if I say (x-a)*(x-b) > 0, I don't know whether the solution is a<x<b or b<x<a.

is there a way to arrive at this?


That's not correct. (x-3)*(x-6)>0 means that x<3 or x>6:
Attachment:
MSP13131f9ac318511bi708000038f28af4a7ie7i53.gif
MSP13131f9ac318511bi708000038f28af4a7ie7i53.gif [ 7.11 KiB | Viewed 588 times ]


Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.

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Re: Doubt on Solution to Inequalities [#permalink] New post 24 Jul 2013, 21:55
Hi Bunuel

Thanks a lot for those links and your wonderful contribution.

I realized that I am struggling specifically with the below:

For example, let us take the equation (x-1)(x-4)=0

So, when I look at the expression (x-1)(x-4), I know that it is a parabola passing through x=1 and x=4.

But, I don't know whether this is V-shaped parabola or an inverted V-shaped parabola. Currently, the way I am doing this is to substitute a value between the roots (for eg. x=2 in the above case) and if it is negative, I know that the curve is a V-shaped parabola. Hence the curve is +ve outside the points x=1 and x=4 and is -ve between the roots.

So, for the inequality (x-1)(x-4)>0 the solution is 4<x<1 (where is the curve +ve)
So, for the inequality (x-1)(x-4)<0 the solution is 1<x<4 (where is the curve -ve)

If you see the above, the crucial thing is to identify whether the curve is V-shaped or inverted V-shaped?

Currently, I am doing it using substitution method, and I feel it is prone to errors. So, I want to know is there any other way apart from substituting values to find out the shape of the curve?

I am extremely sorry for the repeated questions, but I really need some handholding on this topic. Kindly help me on this.
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Re: Doubt on Solution to Inequalities [#permalink] New post 25 Jul 2013, 00:43
Expert's post
gmatter0913 wrote:
Hi Bunuel

Thanks a lot for those links and your wonderful contribution.

I realized that I am struggling specifically with the below:

For example, let us take the equation (x-1)(x-4)=0

So, when I look at the expression (x-1)(x-4), I know that it is a parabola passing through x=1 and x=4.

But, I don't know whether this is V-shaped parabola or an inverted V-shaped parabola. Currently, the way I am doing this is to substitute a value between the roots (for eg. x=2 in the above case) and if it is negative, I know that the curve is a V-shaped parabola. Hence the curve is +ve outside the points x=1 and x=4 and is -ve between the roots.

So, for the inequality (x-1)(x-4)>0 the solution is 4<x<1 (where is the curve +ve)
So, for the inequality (x-1)(x-4)<0 the solution is 1<x<4 (where is the curve -ve)

If you see the above, the crucial thing is to identify whether the curve is V-shaped or inverted V-shaped?

Currently, I am doing it using substitution method, and I feel it is prone to errors. So, I want to know is there any other way apart from substituting values to find out the shape of the curve?

I am extremely sorry for the repeated questions, but I really need some handholding on this topic. Kindly help me on this.


Please go through the links provided in my post above:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Doubt on Solution to Inequalities [#permalink] New post 25 Jul 2013, 07:29
Quote:
Intersection points are the roots of the equation x^2-4x+3=0, which are x_1=1 and x_2=3. "<" sign means in which range of x the graph is below x-axis. Answer is 1<x<3 (between the roots).

If the sign were ">": x^2-4x+3>0. First find the roots (x_1=1 and x_2=3). ">" sign means in which range of x the graph is above x-axis. Answer is x<1 and x>3 (to the left of the smaller root and to the right of the bigger root).


This approach works for any quadratic inequality. For example: -x^2-x+12>0, first rewrite this as x^2+x-12<0 (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).

x^2+x-12<0. Roots are x_1=-4 and x_1=3 --> below ("<") the x-axis is the range for -4<x<3 (between the roots).

Again if it were x^2+x-12>0, then the answer would be x<-4 and x>3 (to the left of the smaller root and to the right of the bigger root).


Yes, this solves my doubt. You're suggesting that whatever the equation is convert it into the form ax^2+bx+c such that the "a" (coeff of x^2) is always +ve. So, for example, if we have to find where is the inverted-V curve +ve, we are actually finding where is the V-curve(mirror image) negative.

My only question is: Can we safely assume that a curve with a +ve coeff of x^2 is always a V-shaped curve?
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Re: Doubt on Solution to Inequalities [#permalink] New post 25 Jul 2013, 09:25
Looks like if a is +ve in ax^2+bx+c=0, then it is a V-shaped curve. Got it from the below link:
http://gmatclub.com/forum/tips-and-tricks-inequalities-150873.html#p1211920
Kindly let me know if I am wrong.
Re: Doubt on Solution to Inequalities   [#permalink] 25 Jul 2013, 09:25
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