Write \(x^2 + 6kx +144\) in the form \((x+p)^2+q\) and thus obtain expressions for \(p\) and \(q\) in terms of \(k\).

We have been asked to complete the square.

In this case, we have \[\begin{align*} x^2 + 6kx +144 &= x^2 + 6kx +9k^2-9k^2 +144\\ &= (x+3k)^2-9k^2 +144.\\ \end{align*}\]

Comparing this expression with \((x+p)^2+q\), we find

\[\begin{align*} p&=3k \\ \text{and}\quad q&=-9k^2+144. \end{align*}\]

Hence find the range of values for \(k^2\) such that \(x^2+6kx+144\) is positive for all values of \(x\), and deduce the corresponding range of values for \(k\).

The smallest value a quadratic of the form \((x+p)^2+q\) takes is \(q\), since \((x+p)^2\ge0\) for all \(p\). This minimum happens when \(x=-p\), which is when \((x+p)^2=0\).

Thus for the quadratic to be positive everywhere, we require \(q>0\). Using the expression for \(q\) obtained above, we have \[-9k^2+144>0\] which rearranges to \[k^2<16.\] Hence \(-4 < k < 4\), which can be also be written as \(\big |k\big |<4\).