Question

How do you solve `16-x^2>0` using a sign chart?

Answer 1

The solution is `x in (-4,+4)`

Explanation:

We need

`a^2-b^2=(a+b)(a-b)`

We factorise the inequality

`16-x^2>0`

`(4+x)(4-x)>0`

Let

`f(x)=(4+x)(4-x)`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-4` `color(white)(aaaa)` `+4` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `4+x` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `4-x` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`

Therefore,

`f(x)>0` when `x in (-4,+4)`