Review question

# For what values of $k$ is $x^2 + 6kx +144$ always positive? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6526

## Solution

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 $\big |k\big |<4$.