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27 Oct 2010, 02:55
Kudos
Bookmarks
Following are 2 sums having both Inequality and Modulus function.
I cannot provide options here, as I picked it from a book giving only the correct answer.
The problem is not in the questions itself, but in the principles that one applies across the 2 problems.
I am not able to grasp a common way of solving them both.
Here it goes:
Q.1> Solve for x,
if |x-2| <= 2 and |x+3| >= 4.
Q.2> Solve for x,
if |x^2 + 3x| + x^2 - 2 >= 0.
I'll confirm and provide the correct answers as soon somebody explains it.
Cheers
R J
I cannot provide options here, as I picked it from a book giving only the correct answer.
The problem is not in the questions itself, but in the principles that one applies across the 2 problems.
I am not able to grasp a common way of solving them both.
Here it goes:
Q.1> Solve for x,
if |x-2| <= 2 and |x+3| >= 4.
Q.2> Solve for x,
if |x^2 + 3x| + x^2 - 2 >= 0.
I'll confirm and provide the correct answers as soon somebody explains it.
Cheers
R J
Archived Topic
Hi there,
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27 Oct 2010, 07:59
Kudos
Bookmarks
Q1:
----
|x-2| <= 2 and |x+3| >= 4.
boundary point 2 and -3
so we need to check for 3 ranges:
1) when X >=2
2) when -3 < X < 2
3) when X < -3
from 1: |x-2| = X-2 < = 2 => X <= 4 . So That means X = 2,3,4
|x+3| = x+3 >=4, So x >=1 . So as we have assumed X >=2, the soultion is x = 2,3,4...
Combining both : That means X = 2,3,4 --- A
from 2: |x-2| = 2-x < = 2 => 0 <= X . So That means X = 0,1 as X < 2
|x+3| = x+3 >=4,So x >=1 . So as we have assumed X < 2, the soultion is x = 1
Combining both : That means X = 1 --- B
from 3: |x-2| = 2-x < = 2 => 0 <= X . So That means X = 0,1 . This is false as we assumed X < -3
So this value is discarded.
So from 1 and 2, the values of x are: 1,2,3,4
(give me a kudo if you like it)
----
|x-2| <= 2 and |x+3| >= 4.
boundary point 2 and -3
so we need to check for 3 ranges:
1) when X >=2
2) when -3 < X < 2
3) when X < -3
from 1: |x-2| = X-2 < = 2 => X <= 4 . So That means X = 2,3,4
|x+3| = x+3 >=4, So x >=1 . So as we have assumed X >=2, the soultion is x = 2,3,4...
Combining both : That means X = 2,3,4 --- A
from 2: |x-2| = 2-x < = 2 => 0 <= X . So That means X = 0,1 as X < 2
|x+3| = x+3 >=4,So x >=1 . So as we have assumed X < 2, the soultion is x = 1
Combining both : That means X = 1 --- B
from 3: |x-2| = 2-x < = 2 => 0 <= X . So That means X = 0,1 . This is false as we assumed X < -3
So this value is discarded.
So from 1 and 2, the values of x are: 1,2,3,4
(give me a kudo if you like it)
