SOLUTION: Please help me solve this: (last time i typed wrong) The quadratic equation {{ (b-c)x^2+(c-a)x+(a-b)=0 }}} has a repeated real solution. Prove that {{{ b=(a+c)/2 }}} Thanks!

Consider    (using capital letters to distinguish the coefficients from the given problem).



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



     



The roots are real and repeated when   or we can say when    (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:



       



This can be expanded, then simplified:

       

       

       



divide thru by 4:

       



       



Notice this is a perfect square:

        



Which has the solution:

        



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.  

 

Notice that  (b-c) + (c-a) + (a-b) = 0.    (It is obvious !)


It means that x= 1 is the root to this quadratic.



But the problem states that the root is REPEATED (!)

It means that the quadratic has the minimum (or the maximum) at the point  x= 1.


In any case, it means that  the point (1,0) in the coordinate plane is the VERTEX of the quadratic function/(parabola).



It implies that  

    1 = -,    (1)    (well known formula for the parabola's symmetry axis)

where "A" is the coefficient at x^2 and "B" is the coefficient at x.



In our case  A = b-c  and  B = c-a;  hence, (1) means that 


    1 = -,    or

    2*(b-c) = -(c-a),

    2b - 2c = -c + a

    2b      = -c + a + 2c = a + c

     b                    =