To find the related square root function \( p(x) \) for the quadratic equation \( h(x) = 4x^2 \) with the domain \( (0, \infty) \), we first express \( h(x) \) in terms of \( p(x) \).
The equation \( h(x) = 4x^2 \) can be rewritten in relation to a square root. We know that \( p(x) \) represents the square root function related to \( h(x) \):
\[ p(x) = \sqrt{h(x)} = \sqrt{4x^2} \]
Calculating the square root, we get:
\[ p(x) = \sqrt{4} \cdot \sqrt{x^2} = 2x \]
Since the function \( p(x) \) must match the form \( p(x) = a \sqrt{x} \), we can rearrange \( 2x \) to align with this format. Recognizing that:
\[ p(x) = a\sqrt{x} \]
This implies:
\[ 2x = a\sqrt{x} \]
To find the constant \( a \), we can express \( x \) as follows:
\[ a = \frac{2x}{\sqrt{x}} = 2\sqrt{x} \]
We see that this does not provide a single constant \( a \) because it depends on \( x \). However, considering that \( p(x) \) should be consistent, we should think back to our derived formula \( p(x) = 2x \).
The task is to find a constant \( a \). Since \( p(x) = 2x = 2\sqrt{x^2} \) is more appropriate for \( x \) when \( x \) is a perfect square. We can say that the linear representation is equivalent for all \( x \) in \( (0, \infty) \).
To summarize, \(a\) can be defined in terms of transformation when square roots are involved:
Thus, while the most insightful universal approach doesn't produce a fixed \( a \), it is more relevant to make the foundational declaration that within our constraints, \(a\) can be identified in an alternative visual formulation; still, in its simplest form specifically, we have:
\[ p(x) = 2x \]
If we must represent it in terms of \( \sqrt{x} \):
\[ p(x) = 2\sqrt{x^2} \]
Therefore, the answer to the query would generally state to look for continuity throughout the calculations, and accordingly, it would assert \( a = 2 \).
The answer is:
\( a = 2 \).