Find the roots of following equation1/x-1/x-2=3, x=0, 2
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We have
(1/x) - [1/(x-2)] = 3
1 1
--- - ------ = 3
x x-2
The common denominator is x(x-2)
1 x-2 1 x
---• ----- - ----- • --- = 3
x x-2 x-2 x
x-2 x
------- - ------- = 3
x(x-2) x(x-2)
x-2 - x
-------- = 3
x(x-2)
-2 = 3x(x-2)
-2 = 3x2-6x
3x2-6x+2 = 0
Using the quadratic formula we can solve for x
-b±√(b2-4ac)
--------------
2a
6±√[62-4(3)(2)]
-------------------
2(3)
6±√(36-24)
-------------
6
6±√12
-------
6
6±2√3
-------
6
3±√3
------
3
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