SOLUTION: 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.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 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.      Log On


   

Question 1209732: 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.
Found 2 solutions by ikleyn, mccravyedwin:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
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.
~~~~~~~~~~~~~~~~~~~~


        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. 

          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.

        Below is my solution for this edited formulation. 


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 =  %282%2Asqrt%28b%5E2-4ac%29%29%2F2 = sqrt%28b%5E2-4ac%29.


Therefore,  (r-s)^2 = b%5E2-4ac = %28-2m%29%5E2+-+4%2A1%2A%28m%5E2-6m%2B11%29 = 

                    = 4m%5E2+-+4m%5E2+%2B+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.


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

Solved.



Answer by mccravyedwin(406) About Me  (Show Source):
You can put this solution on YOUR website!
The answer will be 0, the smallest possible value of (r-s)2

if and only if we can find m such that the vertex of the parabola

f%28x%29=x%5E2+-+2mx+%2B+%28m%5E2+-+6m+%2B+11%29+

is on the x-axis, where both x-intercepts, r and s, the zeros of f(x),
coincide and have difference 0.

We use the vertex formula

%28matrix%281%2C3%2C+-b%2F%282a%29%2C+%22%2C%22%2C+f%28-b%2F%282a%29%29%5E%22%22%29%29%29

-b%2F%282a%29+=+-%28-2m%29%2F%282%281%29%29=m

So we need for the vertex to be the point (m, 0)

f%28-b%2F%282a%29%29+=+m%5E2+-+2m%28m%29+%2B+%28m%5E2+-+6m+%2B+11%29+=+-6m%2B11

The vertex will be on the x-axis if and only if 

-6m + 11 = 0 or m = 11/6.

Since this is possible, the answer is 0.

Edwin