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oops
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what are the values of x that satisfy the ineq x + 1/x < 07 Jul 2007, 15:45
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what are the values of x that satisfy the ineq x + 1/x < 2.5?

not sure of given answer (will post later).



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what are the values of x that satisfy the ineq x + 1/x < 07 Jul 2007, 16:17
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what are the values of x that satisfy the ineq x + 1/x < 2.5?

not sure of given answer (will post later).
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i got: x is grate than 0.5 and less than < 0.
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what are the values of x that satisfy the ineq x + 1/x < 07 Jul 2007, 22:12
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what are the values of x that satisfy the ineq x + 1/x < 2.5?

not sure of given answer (will post later).
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I am getting: 0.5 < x < 2
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what are the values of x that satisfy the ineq x + 1/x < 07 Jul 2007, 23:51
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what are the values of x that satisfy the ineq x + 1/x < 2.5?

not sure of given answer (will post later).

I am getting: 0.5 < x < 2
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x + 1/x < 2.5
=> x^2 + 1 < 2.5x
=> x^2 - 2.5x + 1 < 0
=> (x - 2)*(x - 0.5) < 0

Either (x-2) or (x-0.5) should be less than zero, but not both.
For x>2, both terms are positive. For x<0.5, both terms are negative, hence, the product is positive.

So, the range of x satisfying the inequality should be 0.5 < x < 2.
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what are the values of x that satisfy the ineq x + 1/x < 08 Jul 2007, 00:04
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what are the values of x that satisfy the ineq x + 1/x < 2.5?

not sure of given answer (will post later).

I am getting: 0.5 < x < 2

x + 1/x <2> [size=18]x^2 + 1 <2> x^2 - 2.5x + 1 <0> (x - 2)*(x - 0.5) <0>2, both terms are positive. For x<0.5, both terms are negative, hence, the product is positive.

So, the range of x satisfying the inequality should be 0.5 < x < 2.
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what if x < 0 ?

:?
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what are the values of x that satisfy the ineq x + 1/x < 08 Jul 2007, 00:11
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oops
what are the values of x that satisfy the ineq x + 1/x < 2.5?

not sure of given answer (will post later).

I am getting: 0.5 < x < 2

x + 1/x <2> [size=18]x^2 + 1 <2> x^2 - 2.5x + 1 <0> (x - 2)*(x - 0.5) <0>2, both terms are positive. For x<0.5, both terms are negative, hence, the product is positive.

So, the range of x satisfying the inequality should be 0.5 < x < 2.

what if x < 0 ?

:?
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You are right. x < 0 does satisfy the inequality. But what is wrong with my approach. :?
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what are the values of x that satisfy the ineq x + 1/x < Updated on: 08 Jul 2007, 00:47
Originally posted by KillerSquirrel on 08 Jul 2007, 00:41.
Last edited by KillerSquirrel on 08 Jul 2007, 00:47, edited 1 time in total.
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sumande
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oops
what are the values of x that satisfy the ineq x + 1/x < 2.5?

not sure of given answer (will post later).

I am getting: 0.5 < x < 2

x + 1/x <2> [size=18]x^2 + 1 <2> x^2 - 2.5x + 1 <0> (x - 2)*(x - 0.5) <0>2, both terms are positive. For x<0.5, both terms are negative, hence, the product is positive.

So, the range of x satisfying the inequality should be 0.5 < x < 2.

what if x < 0 ?

:?

You are right. x < 0 does satisfy the inequality. But what is wrong with my approach. :?
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when you multply by x an inequality you need to be extra cerful. multiplying by a negative number will result in "sign flip" - hence whenever x < 0:

x + 1/x < 2.5

x^2 +1 > 2.5x

:)
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what are the values of x that satisfy the ineq x + 1/x < 08 Jul 2007, 00:44
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when you multply by x an inequality you need to be extra cerful. multiplying by a negative number will result in "sign flip" - hence whenever x < 0:

x + 1/x <2>[/b] 2.5x

:)
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Of course !! My stupidity. :oops:
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what are the values of x that satisfy the ineq x + 1/x < Updated on: 08 Jul 2007, 12:04
Originally posted by oops on 08 Jul 2007, 11:54.
Last edited by oops on 08 Jul 2007, 12:04, edited 1 time in total.
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format messed up due to HTML and inequality sign incompatibility - hence deleted
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what are the values of x that satisfy the ineq x + 1/x < 08 Jul 2007, 11:56
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forgot to disable HTML - sorry - ignore previous post - reposting

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sumande
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oops
what are the values of x that satisfy the ineq x + 1/x < 2.5?

not sure of given answer (will post later).

I am getting: 0.5 < x < 2

x + 1/x <2> [size=18]x^2 + 1 <2> x^2 - 2.5x + 1 <0> (x - 2)*(x - 0.5) <0>2, both terms are positive. For x<0.5, both terms are negative, hence, the product is positive.

So, the range of x satisfying the inequality should be 0.5 < x < 2.

what if x < 0 ?

:?

You are right. x < 0 does satisfy the inequality. But what is wrong with my approach. :?

when you multply by x an inequality you need to be extra cerful. multiplying by a negative number will result in "sign flip" - hence whenever x < 0:

x + 1/x < 2.5

x^2 +1 > 2.5x

:)
Show more



am glad you brought this up - is the same problem i had with the given solution. hence the reason i posted this question. the given solution actually flips the sign but still seems to give the wrong answer:

x + 1/x < 25/10
=> (x^2 + 1)/x < 25/10

case 1: x > 0:
10x^2 + 10 < 25x
=> (10x-5)(x-2) < 0
=> 1/2 < x < 2

case 2: x < 0:
10x^2 + 10 > 25x
=> (10x-5)(x-2) > 0
=> but x between 1/2 and 2 is not permissible since in this case, we are assuming x < 0 to begin with.

Hence the soln concludes that x is between 1/2 and 2, which seems incorrect, since if you simply inspect the given ineq x + 1/x < 2.5, any value of x that is -ve will satisfy the inequation. The solution set becomes: {x< 0} U {1/2<x<2}

My sense is that you have to consider 0 as another "pivotal" point since the inequation is (10x^2 + 10)/x > 25 (i.e. it is divided by x which if 0, will make it undefined) and inspect both sides of 0 as well. This is as far as my knowhow goes. For more on this topic, please search for postings by "Fig", the absolute-inequation expert and also "hobbit" (Fig or hobbit, if you happen to read this, please confirm/clarify. Thx)
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what are the values of x that satisfy the ineq x + 1/x < 08 Jul 2007, 12:56
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x + 1/x <25> (x^2 + 1)/x <25> 0:
10x^2 + 10 <25x> (10x-5)(x-2) <0> 1/2 < x < 2

case 2: x <0> 25x
=> (10x-5)(x-2) > 0
=> but x between 1/2 and 2 is not permissible since in this case, we are assuming x < 0 to begin with.
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thats an important point to remember when working on inequalities..

but i couldnot follow the following as under:
Quote:
Hence the soln concludes that x is between 1/2 and 2
Show more
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what are the values of x that satisfy the ineq x + 1/x < 08 Jul 2007, 13:11
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oops
x + 1/x (x^2 + 1)/x 0:
10x^2 + 10 (10x-5)(x-2) 1/2 25x
=> (10x-5)(x-2) > 0
=> but x between 1/2 and 2 is not permissible since in this case, we are assuming x < 0 to begin with.

thats an important point to remember when working on inequalities..

but i couldnot follow what you said under:
oops
Hence the soln concludes that x is between 1/2 and 2
Show more



Sorry if not clear - let me rephrase:

The official solution says that the final answer is: 1/2 < x < 2
It seems incorrect to me since it excludes all x < 0
The answer I get is {x < 0} U { 1/2 < x < 2}
(i.e. we have to check for validity in a total of four domains)

Fig/Hobbit/Other experts, plz comment.



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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