Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
_________________

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

RELATED VIDEO

_________________

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.