Question 1209732
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Suppose r and s are the values of x that satisfy the equation
x^2 - 2mx + (m^2 - 6m + 11) = 0
for some real number m.  Find the minimum real value of (r - s)^2.
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        As I read a problem,  I see that it is written unprofessionally.

        Although from the great distance it looks  OK,  but in reality it should be RE-WRITTEN.


        My editing is below.


<pre>
          Suppose r and s are real roots of the equation
              x^2 - 2mx + (m^2 - 6m + 11) = 0
          for some real number m.  Find the minimum real value of (r - s)^2.
</pre>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Below is my solution for this edited formulation.



<pre>
If you use the quadratic formula for the roots, you will see that the difference between the roots 

is the square root of the discriminant


    the difference between the roots =  {{{(2*sqrt(b^2-4ac))/2}}} = {{{sqrt(b^2-4ac)}}}.


Therefore,  (r-s)^2 = {{{b^2-4ac}}} = {{{(-2m)^2 - 4*1*(m^2-6m+11)}}} = 

                    = {{{4m^2 - 4m^2 + 24m- 44}}} = 24m - 44.


The roots are real numbers if and only if the discriminant  (24m - 44)  is non-negative
Otherwise, the roots are complex non-real numbers.


As we consider the case when the roots r and s are real numbers,
we must assume that  the number  24-44 is non-negative.


Then the difference (r-s)^2 is minimal, when the discriminant 24m-44 is zero.

It happens at m = 44/24 = 11/6.


<U>ANSWER</U>.  The minimum of  (r-s)^2, assuming r and s are real roots, is 0 (zero) at m = 11/6.
</pre>

Solved.