Solve the following ? Please

Question

#|x^2-4x+3|=x+1#

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Answer 1 · 2018-06-17T06:12:39

#x=(5+-sqrt17)/2#

If #|x^2-4x+3|=x+1#, we have #x>=-1# as #|x^2-4x+3|>=0#. This gives us the domain of #x#.

Now we have two possibilities

#x^2-4x+3=x+1# or #x^2-5x+2=0# i.e.

#x=(5+-sqrt(5^2-8))/2=(5+-sqrt17)/2#

Observe that both are in domain.

If #x^2-4x+3=-x-1# or #x^2-3x+4=0#. But here discriminant is negative #3^2-4×1×4=-7#. Hence we do not have any solution.