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.

Thus the answer is (a).