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Question
Solve the following quadratic equations by factorization:
`1/((x-1)(x-2))+1/((x-2)(x-3))+1/((x-3)(x-4))=1/6`
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Solution
We have been given,
`1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)=1/6`
`((x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2))/((x-1)(x-4)(x-2)(x-3))=1/6`
`(3(x^2-5x+6))/((x^2-5x+4)(x^2-5x+6))=1/6`
18 = x2 - 5x + 4
x2 - 5x - 14 = 0
x2 - 7x + 2x - 14 = 0
x(x - 7) + 2(x - 7) = 0
(x + 2)(x - 7) = 0
Therefore,
x + 2 = 0
x = -2
or,
x - 7 = 0
x = 7
Hence, x = -2 or x = 7.
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