Review question

# When is the inequality $x^4<8x^2+9$ satisfied? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9603

## Solution

The inequality $x^4<8x^2+9$ is satisfied precisely when $\text{(a)}-3<x<3 \quad \text{(b)} \quad 0<x<4 \quad \text{(c)} \quad 1<x<3 \quad \text{(d)}-1<x<9 \quad \text{(e)}-3<x<-1.$

We have \begin{align*} x^4<8x^2+9 \\ \iff x^4-8x^2-9&<0 \\ \iff (x^2+1)(x^2-9)&<0 \\ \iff (x^2+1)(x+3)(x-3)&<0. \end{align*}

Now $x^2 + 1$ is always positive, so we are interested in when $(x+3)(x-3)$ is negative.

When $x < -3$, both brackets are negative, and when $x > 3$, both brackets are positive.

In between these values, one bracket is negative and one is positive, and so $(x+3)(x-3)$ is negative.