Question

What is the length of the segment of the number line consisting of the points that satisfy `(x-4)^2 \le 9`?

I thought it was `7` but that was wrong...
Can't `x` be `1, 2, 3, 4, 5, 6,` and `7`?

*This question is from AoPS (AoPS info below).
Subject: Algebra
Focus: Quadratic Inequalities

Answer 1

6

Explanation:

OHHHH OKAY SO I'M DUMB. I got it wrong because it's asking for the length, and even though there are 7 numbers, the distance is 6.

On to the Real Explanation

First, take the square root of both sides. Then you get:
`x-4\le3`
Add `4` to both sides.
`x\le7`
However, if you think about it (and look at what the question is asking), `x` cannot possibly equal all of the values less than `7`.
Checking different values, you can see that 0 doesn't work.
And so,
`x` can be anywhere from `1` to `7`.
Not a very good solution, I know, but...
oh! here's

AoPS' Solution:

Since the square of `x-4` is at most 9, the value of `x-4` must be between `-3` and `3` (or equal to either). So, we have `-3 \le x-4 \le 3`. Thus, `1 \le x \le 7`. Therefore, our answer is `6`.

OR -

If `(x-4)^2 \le 9`, then `x` can be no more than 3 away from 4. Therefore, the values of `x` from 1 to 7 satisfy the inequality, and our answer in `6`.