Question

Find x in terms of a, b and c:

(a/x-a) +(b/x-b)=(2c/x-c), x is not equal to a, b, c

Answer 1

Hey!

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We have to find the roots right?

Refer to the pic.

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Hope it helps...!!!

Answer 2

yeah sure for your help

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Thanks !

Here is your answer which you are searching for

It is given that

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

$\frac{[ a ( x-b ) + b (x-a) ]}{(x-a) (x-b) = \frac{2c}{x-c}}$

( x-c )[a(x-b)+b(x-a)] = 2c (x-a) (x-b)$ax^{2} -2abx +bx^{2} -acx + 2abc -bcx = 2cx^{2} - 2bcx - 2acx + 2abc$

$ax^{2} +bx^{2} -2cx^{2} = 2abx -acx - bcx$

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

( a + b- 2c)$x^{2}$ - x( 2ab -ac - bc ) = 0

x[x[( a + b - 2c) x ( 2ab -ac -bc )] = 0

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

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

x =0 or $\frac{ ( 2ab -ac -bc ) }{(a + b -2c)}$

Thus , the two roots of the given equation are x = 0 or x = $\frac{ ( 2ab -ac -bc ) }{(a + b -2c)}$

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