Question

How do you solve `x^2-6x+8>=0` using a sign chart?

Answer 1

I'd never heard of a "sign chart". We didn't use them back when I was in school, so many many years ago. I had to look this up.

First thing is to factor this puppy:

`x^2 - 6x + 8 = (x-2) (x-4)`

So your x-axis intercepts are at x = 2 and x = 4.

Our original equality is `>=`, and `f(x)` is `x^2 - 6x + 8`, and critical points are 2 and 4. Imagine these marked on a number line.

We build a chart starting with a number to the left of 2, then 2, then a number between 2 and 4, then 4, then a number greater than 4.

when x is 1, f(x) = 3. POSITIVE.
when x is 2, f(x) = 0.
when x is 3, f(x) = -1. NEGATIVE.
when x is 4, f(x) = 0.
when x is 5, f(x) = 3. POSITIVE.

So this would be your chart, I think. Hope I did that right. Your teacher might want it formatted in a different way. It seems like too much work to me. But anyway, from the chart, you can see that the values of x that meet the original inequality are all x where x <= 2 and x >= 4.