Find the roots of the equation ( b-c )x 2 + ( c-a )x + (a-b) = 0
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(b-c)x² + (c-a)x + (a-b) =0
(b-c)x² + (c-b)x + (b-a)x + (a-b) =0
(b-c)x² +[-(b-c)x] +(b-a)x + [-(b-a)] =0
(b-c)x [x-1] +(b-a)[x-1] =0
(x-1) [(b-c)x + (b-a)] =0
(x-1) =0 or (b-c)x + (b-a) =0
x=1 or (b-c)x = -(b-a)
(b-c)x = (a-b)
x = (a-b)/(b-c)
Therefore x= 1 x= (a-b)/(b-c)
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