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# some maths doubts...

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Manager
Joined: 03 Jul 2006
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10 Jul 2006, 07:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can someone please explain these ?

1) x^2 > 2

how can we get the following solution ?
x < -sqrt(2) < -1 or x > sqrt(2) > 1

2) x^2 > x

how to derive the solution for x ?

3) What does the following mean for a triangle

AC^2+AB^2>BC^2

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Joined: 05 Apr 2005
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10 Jul 2006, 09:34
rkatl wrote:
1) x^2 > 2
how can we get the following solution ?
x < - sqrt(2) < -1 or x > sqrt(2) > 1

x^2 > 2
take the sqrt
x > sqrt(2) or x < - sqrt(2). and - sqrt(2) is less than -1.
take any value. lets say x^2 = 25. x can either be 5 or smaller than -5.
x (5) > sqrt (2).
x (-5) < - sqrt (2).

rkatl wrote:
2) x^2 > x
how to derive the solution for x ?

x^2 > x
x^2 - x > 0
x (x- 1) > 0
here it ios correct to say x > 1 but not x>0. actually x>1 but is <0.

rkatl wrote:
3) What does the following mean for a triangle
AC^2 + AB^2 > BC^2

this means the triangle is not a right triangle.
Re: some maths doubts...   [#permalink] 10 Jul 2006, 09:34
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