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Solving an inequality equation

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by Vemuri » Wed Apr 08, 2009 8:21 pm
x² - 6x + 8 > 0 Solve for x.

I understand that for the above inequality, the 2 factors are (x-4) & (x-2). So, in effect (x-4) * (x-2) > 0. I would interpret the solution to be x>2 or x>4. But, because the inequality has no = sign, x=4 will not be part of the solution & so, x>2 will not be valid. But by saying that x>4, I would be missing x=3 (which is surely a solution).

As you can see I am completely tangled trying to find the solution. Can anyone please help?
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by gmat740 » Wed Apr 08, 2009 9:04 pm
Some fundamental rules of Inequality
Case I

(x-a)(x-b)>0

where a>b

so,

x>a and x<b

Case II

(x-a)(x-b) < 0

so b<x< a
provided a>b


so the question you posted is Case I


(x-4)(x-2) >0
so either x> 4 or x<2

had it been:

(x-4)(x-2)<0

Case II



2 < x < 4


Hope this Helps.Let me know you have some more questions.


Karan
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by Vemuri » Thu Apr 09, 2009 1:33 am
Thank you Karan for sharing the rules. I want to go a step further in understanding the rules. I will appreciate if you can elaborate for a clearer understanding.

When a>b

Case I

(x-a)(x-b)>0

x>a and x<b --> How did we determine this?


Case II

(x-a)(x-b) < 0

b<x< a --> How did we determine this?
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by lav » Thu Apr 09, 2009 3:16 am
vemuri if you put x = 3 in x² - 6x + 8 or (x-4)(x-2)
you get

(3-4)(3-2) = (-1)(1) = -1

which is less that 0
hence x=3 is not a soln for x² - 6x + 8 >0 because -1<0 not greater
Kid in Verbal :(
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by gmat740 » Thu Apr 09, 2009 3:55 am
Case I

(x-a)(x-b)>0

x>a and x<b --> How did we determine this?


Case II

(x-a)(x-b) < 0

b<x< a --> How did we determine this?

this is something which comes from the Union and set theory, when you plot x, a and b on a number line;it will be clear
Case I

(x-a)(x-b)>0

x>a and x<b --> How did we determine this?
this means x belongs to a set of numbers: (Union of numbers which are greater that a and numbers which are less than b) - (numbers which fall between a and b)

-infinity -----------------b---------0----------a-----------------------+infinity
Case I(Bold Face part)


Similarly
Case II:

-infinity -----------------b---------0-------a--------------------+infinity

Hope this helps

karan





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by Vemuri » Thu Apr 09, 2009 7:29 am
lav wrote:vemuri if you put x = 3 in x² - 6x + 8 or (x-4)(x-2)
you get

(3-4)(3-2) = (-1)(1) = -1

which is less that 0
hence x=3 is not a soln for x² - 6x + 8 >0 because -1<0 not greater
Yep, you are right. I got tangled so badly I was not checking my observations. Thanks for correcting.
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by Vemuri » Thu Apr 09, 2009 7:32 am
gmat740 wrote:
Case I

(x-a)(x-b)>0

x>a and x<b --> How did we determine this?


Case II

(x-a)(x-b) < 0

b<x< a --> How did we determine this?

this is something which comes from the Union and set theory, when you plot x, a and b on a number line;it will be clear
Case I

(x-a)(x-b)>0

x>a and x<b --> How did we determine this?
this means x belongs to a set of numbers: (Union of numbers which are greater that a and numbers which are less than b) - (numbers which fall between a and b)

-infinity -----------------b---------0----------a-----------------------+infinity
Case I(Bold Face part)


Similarly
Case II:

-infinity -----------------b---------0-------a--------------------+infinity

Hope this helps

karan

Thanks for being patient Karan. Much appreciated.
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by gmat740 » Thu Apr 09, 2009 7:49 am
You are welcome

We are prospective B-Schools candidates and we must learn to work in group

This is just another dimension to it

Regards

Karan
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by aj5105 » Sun May 03, 2009 8:34 pm
Please note the correction.

Check Ian's post for more details.

https://www.beatthegmat.com/inequality-p ... 12734.html

gmat740 wrote:Some fundamental rules of Inequality
Case I

(x-a)(x-b)>0

where a>b

so,

x>a OR x<b

Case II

(x-a)(x-b) < 0

so b<x< a
provided a>b


so the question you posted is Case I


(x-4)(x-2) >0

so either x> 4 or x<2

had it been:

(x-4)(x-2)<0

Case II



2 < x < 4


Hope this Helps.Let me know you have some more questions.


Karan
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