Question 1161183
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Consider  {{{Ax^2 + Bx + C = 0 }}}  (using capital letters to distinguish the coefficients from the given problem).

If one applies the quadratic formula, they get the roots:

     {{{ x = (-B +- sqrt(B^2 - 4AC))/(2A) }}}

The roots are real and repeated when  {{{ sqrt(B^2-4AC) = 0 }}} or we can say when {{{ B^2 - 4AC = 0}}}   (this is called the discriminant).


Compare A,B,C to the given problem:

  A = b-c
  B = c-a
  C = a-b

Now use the right hand sides of these to get the discriminant:

       {{{ (c-a)^2 - 4(b-c)(a-b) = 0 }}}

This can be expanded, then simplified:
       {{{ c^2-2ac+a^2  -4ab+4b^2+4ac-4bc = 0 }}}
       {{{ c^2+2ac+a^2  +4b^2 - 4ab-4bc  = 0 }}}
       {{{ 4b^2 - 4(a+c)b + (a+c)^2 = 0 }}}

divide thru by 4:
       {{{ b^2 - (a+c)b + (a+c)^2/4 = 0 }}}

       {{{ b^2 - (a+c)b + ((a+c)/2)^2 =  0 }}}

Notice this is a perfect square:
        {{{ (b - (a+c)/2)^2 = 0 }}}

Which has the solution:
        {{{ b = (a+c)/2 }}}

Recall:
It is this relationship between a,b,and c that must hold in order for the original descriminant to be zero.   That in turn, implies repeated real roots.