Question

complete the square (1/4x^2)+6x+7

Answer 1

That's \(\dfrac{1}{4}x^2\), and not \(\dfrac{1}{4x^2}\), correct?

Answer 2

\[\begin{align*}\frac{1}{4}x^2+6x+7&=\frac{1}{4}\left(x^2+24x+\frac{7}{4}\right)\\
&=\frac{1}{4}\left(x^2+24x+144-144+\frac{7}{4}\right)\\
&=\frac{1}{4}\left((x+12)^2+\cdots\right)
\end{align*}\]