#### 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(x^{2}-(b + c)x + bc) + b(x^{2} - (a + c)x + ac) = 2c(x^{2} - (a + b)x + ab)

(a + b - 2c)x^{2} - (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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#### APPEARS IN

Solution Solve the Following Quadratic Equations by Factorization: `A/(X-a)+B/(X-b)=(2c)/(X-c)` Concept: Solutions of Quadratic Equations by Factorization.