Tricky!!!

Start with: Average \(= \frac{(x-1)^2 + (x+3)^2}{2}\)

Expand numerator: Average \(= \frac{(x^2 - 2x + 1) + (x^2 + 6x + 9)}{2}\)

Simplify numerator: Average \(= \frac{2x^2 + 4x + 10}{2}\)

Simplify numerator: Average \(= x^2 + 2x + 5\)

This is where it gets tricky!!!
You may notice that we kind of have a perfect square hiding in our algebraic expression.
We know that \(= x^2 + 2x + 1\) is a perfect square because we can rewrite it as \((x + 1)^2\)

So let's take: \(= x^2 + 2x + 5\)
And rewrite it as: \(= x^2 + 2x + 1 + 4\)
Now factor the first part to get: \(= (x + 1)^2 + 4\)

So, the average \(= (x + 1)^2 + 4\)

Since \((x + 1)^2 ≥ 0\) for all values of x, we can conclude that \((x + 1)^2 + 4 ≥ 4\) for all values of x

In other words:
QUANTITY A: Some number greater than or equal to 4
QUANTITY B: 3

Answer: A

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