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Re: What are the values of x that satisfy [#permalink]
When x^2-x+1 = -(2x-1) we assume that x^2-x+1<0 that is why we put a negative sign for opening the mod function.
On solving above equations we get x=0,-1 on putting in equation x^2-x+1 we get 1 for x=0 and 3 for x=-1 both >0 so we have to discard those answers.
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Re: What are the values of x that satisfy [#permalink]
|x^2-x+1| = 2x-1
since, x^2-x+1 = (x-1/2)^2 +3/4 >0 => |x^2-x+1| = x^2-x+1
Hence |x^2-x+1| = x^2-x+1 = 2x-1 => x^2-3x+2 = 0 => (x-1)(x-2) = 0 => x=1, 2, so correct answers are B and D only. A can't be possible.
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Re: What are the values of x that satisfy [#permalink]
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tapas3016 wrote:
If someone is geek like me who loves to solve equations, then there will be 2 equations after removing modulus:-

x^2-x+1 = 2x-1 and x^2-x+1 = -2x+1

Solving these

we will get x=1,-1,2,0

Hence, A B and D.

|x^2-x+1| = 2x-1
since, x^2-x+1 = (x-1/2)^2 +3/4 >0 => |x^2-x+1| = x^2-x+1
Hence |x^2-x+1| = x^2-x+1 = 2x-1 => x^2-3x+2 = 0 => (x-1)(x-2) = 0 => x=1, 2, so correct answers are B and D only. A can't be possible.
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Re: What are the values of x that satisfy [#permalink]
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