2x=|x|+|x-1|
each term on the right >= 0 so LHS >= 0 or x >= 0
hence we get 2x = x + | x-1| or |x-1| = x
as x-1 cannot be x so (x-1) < 0 and hence - (x-1) = x or x = 1/2
it satisfies (x-1) < 0 and so x= 1/2 is the solution
each term on the right >= 0 so LHS >= 0 or x >= 0
hence we get 2x = x + | x-1| or |x-1| = x
as x-1 cannot be x so (x-1) < 0 and hence - (x-1) = x or x = 1/2
it satisfies (x-1) < 0 and so x= 1/2 is the solution
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