Suppose the difference of two numbers is 3. Does the sum of their squares have a minimum or maximum value? What is that value?

it has a minimum. We know that when you square a negative number you get a positive number. Lets start with -1 and 2. -1^2+2^2=1+4=5. Now lets check -2 and 1. -2^2+1^2=4+1=5, which is the same thing. If we check (-3 and 0) and (0 and 3), we find that they are both bigger, and if you keep going lower or higher you will get bigger numbers. So the answer is minimum, with a value of 5.

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I disagree with the answer given above.
Nowhere does it say the numbers have to be integers.

To determine whether the sum of the squares of two numbers has a minimum or maximum value when their difference is 3, let's break down the problem step by step.

Let the two numbers be x and y, where x > y. We know that x - y = 3, which implies x = y + 3.

Now let's consider the sum of their squares, which is S = x^2 + y^2. Substituting x = y + 3, we have:

S = (y + 3)^2 + y^2

Expanding the equation, we get:

S = y^2 + 6y + 9 + y^2

Simplifying further, we find:

S = 2y^2 + 6y + 9

This is a quadratic equation in terms of y. To find the minimum or maximum value of S, we can analyze the quadratic equation.

The coefficient of y^2 is positive, indicating that the parabola opens upwards. Therefore, the sum of the squares (S) will have a minimum value rather than a maximum.

Now, let's determine the minimum value of the quadratic equation. We know that the minimum or maximum occurs when the derivative of the quadratic equation is zero.

Differentiating S with respect to y, we have:

dS/dy = 4y + 6

Setting dS/dy equal to zero and solving for y, we get:

4y + 6 = 0
4y = -6
y = -6/4
y = -3/2

Now, substitute this value back into the quadratic equation to find the minimum value:

S = 2(-3/2)^2 + 6(-3/2) + 9
S = 2(9/4) - 18/2 + 9
S = 9/2 - 18/2 + 9
S = 9/2 - 9 + 9
S = 9/2

Therefore, the sum of the squares of the two numbers has a minimum value of 9/2 when their difference is 3.