Question 1031664
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Find all the real solution of the equation 
x^2-|x-1|-3=0
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{{{x^2 - abs(x-1) - 3}}} = {{{0}}}.        (1)


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1. Consider the domain x >= 1.
   In this domain |x-1| = x-1,
   and yours equation (1) becomes

   {{{x^2 - (x-1) - 3}}} = {{{0}}},  or

   {{{x^2 - x - 2}}} = {{{0}}},

   (x-2)*(x+1) = 0

   with the roots x = 2 and x = -1.

   Of them, only x = 2 belongs to the domain x >= 1.
   So, x = 2 is the ONLY solution of the equation (1) in this domain.


2. Next, consider the domain x < 1.
   In this domain |x-1| = -(x-1). 
   and yours equation (1) becomes

   {{{x^2 - (-(x-1)) - 3}}} = {{{0}}},  or

   {{{x^2 + x - 4}}} = {{{0}}}.

   Find the roots using the quadratic formula.

   {{{x[1]}}} = {{{(-1 + sqrt (17))/2}}},  {{{x[2]}}} = {{{(-1 - sqrt (17))/2}}}.

   Of them, only {{{x[2]}}} = {{{(-1 - sqrt (17))/2}}} belongs to the domain x < 1.

   So, {{{x[2]}}} = {{{(-1 - sqrt (17))/2}}}  is the ONLY solution of the equation (1) in this domain.

<U>Answer</U>.  The solutions are  x = 2  and  x = {{{(-1 - sqrt (17))/2}}}.
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