Question

How do you solve `(x^2-9)*(x^2-16)<=0`?

Answer 1

`f(x) = (x^2-9)(x^2-16)` is a continuous function with zeros at `x=+-4` and `x=+-3` of multiplicity `1`.

Hence solution: `x in [-4, -3] uu [3, 4]`

Explanation:

Let `f(x) = (x^2-9)(x^2-16)`

`= (x-3)(x+3)(x-4)(x+4)`

`f(x)` is a continuous function with zeros at `x=+-4` and `x=+-3`.

Each of the roots of `f(x) = 0` is of multiplicity `1`, hence `f(x)` changes sign at those points.

For large positive and negative values of `x`, `f(x)` is large and positive.

Hence `f(x) <= 0` when `x in [-4, -3] uu [3, 4]`

Here's a graph of `f(x)/20`...

graph{(x^2-9)(x^2-16)/20 [-9.68, 10.32, -1.52, 8.48]}