What method is required to find out "for what values of k is $4x^2+kx+\frac14$ a perfect square?" 
For what values of $k$ is this expression a perfect square?
$$4x^2+kx+\frac14$$

I'm thinking maybe the perfect square trinomial is supposed to be used, as I had to use that formula on the problem prior to this one. However, I don't know how to apply it, since I don't know which of the above terms would be $a^2$, $2ab$ and $b^2$. How can one deduce what term corresponds to what term in the general formula? Or is this not the formula to use at all?
EDIT:
User Exodd showed a general formula that when I followed, gave the same answer as user Jitendra Singh, who answered this question with a different formula.
The general formula is this:
$$(ax+b)^2 = a^2x^2+2abx+b^2$$
Which is a variation on the simpler perfect square trinomial
$$(a+b)^2=a^2+2ab+b^2$$
Now, if one assumes that $4x^2$ is $a^2x^2$, $kx$ is $2abx$ and $\frac14$ is $b^2$, then the calculation is quite simple. $k$ is then equal to $2ab$, which is equal to $\pm 2$, the sign depending on whether the $4$, or $a^2$, is the product of $2 \times 2$ or $-2 \times -2$, which creates:
$$2 \times 2 \times \frac12 = 2$$
or
$$2 \times -2 \times \frac12 = -2$$
However, this is assuming the aforementioned correspondence. It could be that $4x^2$ is $a^2x^2$, $kx$ is $b^2$ and $\frac14$ is $2abx$. In that case:
$$\frac14 = 2 \times 2 \times \sqrt{kx} \times x$$
or
$$\frac14 = 2 \times -2 \times \sqrt{kx} \times x$$
Furthermore, there could be a different correspondence. What if $4x^2$ is $b^2$, $kx$ is $a^2x^2$ and $\frac14$ is $2abx$? There are other correspondences as well. What says that the first one, that gave the answer of $\pm2$, is correct?

We consider a polynomial in $x$
\begin{align*}
p(x)=4x^2+kx+\frac{1}{4}
\end{align*}
which has degree $2$, constant term $\frac{1}{4}$, linear term $kx$ and square term $4x^2$.
We are looking for a representation as perfect square
\begin{align*}
q(x)=(ax+b)^2
\end{align*}
We recall two polynomials are equal, i.e. $p=q$ if they have the same degree and the coefficients of like powers of $x$ are equal.
\begin{align*}
p(x)&=q(x)\\
4x^2+kx+\frac{1}{4}&=a^2x^2+2abx+b^2\tag{1}
\end{align*}
From (1) we see $\deg p = \deg q$ and we obtain by comparison of coefficients
\begin{align*}
4&=a^2\\
k&=2ab\\
\frac{1}{4}&=b^2
\end{align*}
Note: The equality of polynomials is based upon an identity theorem of polynomials which can be found for instance in this paper as theorem 4 and which is nicely answered by @IvesDaoust in this MSE post.
