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Question
Solve the following quadratic equations by factorization:
`1/x-1/(x-2)=3` , x ≠ 0, 2
Solution
We have been given
`1/x-1/(x-2)=3`
-2 = 3x2 - 6x
3x2 - 6x + 2 = 0
`3x^2 - (3 + sqrt3)x-(3-sqrt3)x+3-sqrt3+sqrt3-1=0`
`x(3x-3-sqrt3)+((-3+sqrt3)/3)(3x-3-sqrt3)=0`
`((3x-3+sqrt3)/3)(3x-3-sqrt3)=0`
`(sqrt3x-sqrt3+1)(sqrt3x-sqrt3-1)=0`
Therefore,
`sqrt3x-sqrt3+1=0`
`sqrt3x=sqrt3-1`
`x(sqrt3-1)/sqrt3`
or
`sqrt3x-sqrt3-1=0`
`sqrt3x=sqrt3+1`
`x(sqrt3+1)/sqrt3`
Hence, `x(sqrt3-1)/sqrt3` or `x(sqrt3+1)/sqrt3`
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