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

Question

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?

Answer 1

According to this if $b^2=4ac$ then the expression is a perfect square.

So $$ k^2=4 (4) (\frac{1}{4}) $$

So $$ k=±2 $$

Edit:

See I am going to show you why we assume $4k^2$ as $x^2$ and $4$ as $x$.

So lets compare the $2$ expressions-

$$ a^2x^2+2abx+b^2=4x^2+kx+4 $$

Now see if we consider $4x^2$ as $b^2$ then notice this is wrong because $4x^2$ has $x^2$ sign which means $4x^2$ can only considered with a term having $x^2$. Thus we can say-

$$ a^2x^2=4x^2 , 2abx=kx, b^2=4$$

Now you can compare and find the answer like you did in question

Answer 2

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.