Question

$\frac{a}{x-a}$ + $\frac{b}{x-b}$ = $\frac{2c}{x-c}$

Solve for x

Answer 1

Given,a/(x-a) + b/(x-b) = 2c/(x-c)

[a(x-b)+b(x-a)]/(x-a)(x-b) = 2c/(x-c)

(x-c)[a(x-b)+b(x-a)] = 2c(x-a)(x-b)

ax2 – 2abx + bx2 - acx + 2abc – bcx = 2cx2 – 2bcx – 2acx + 2abc

ax2+ bx2 - 2cx2 = 2abx – acx – bcx

(a+b-2c)x2  = x(2ab – ac – bc)

(a+b-2c)x2 - x(2ab – ac – bc) = 0

x[(a+b-2c)x - (2ab – ac – bc)] = 0

x = 0  or  (a+b-2c)x - (2ab – ac – bc) =0

x = 0  or  (a+b-2c)x = (2ab – ac – bc)

x = 0 or x = (2ab – ac – bc) / (a+b-2c)

thus the  2 roots of the given equation are x = 0 and  x = (2ab – ac – bc) / (a+b-2c)

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