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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.
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 = = .
Therefore, (r-s)^2 = = =
= = 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.
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
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
So we need for the vertex to be the point (m, 0)
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