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Question
Solve the following quadratic equations by factorization:
`a/(x-a)+b/(x-b)=(2c)/(x-c)`
Solution
We have been given,
`a/(x-a)+b/(x-b)=(2c)/(x-c)`
a(x - b)(x - c) + b(x - a)(x - c) = 2c(x - a)(x - b)
a(x2-(b + c)x + bc) + b(x2 - (a + c)x + ac) = 2c(x2 - (a + b)x + ab)
(a + b - 2c)x2 - (2ab - ac - bc)x = 0
x[(a + b - 2c)x - (2ab - ac - bc)] = 0
Therefore,
x = 0
Or,
(a + b - 2c)x - (2ab - ac - bc) = 0
(a + b - 2c)x = (2ab - ac - bc)
`x=(2ab - ac - bc)/(a + b - 2c)`
Hence, x = 0 or `x=(2ab - ac - bc)/(a + b - 2c)`
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