Four. (This question is worth 20 points)
Let the function $f(x)=a x^{2}+8 x+3(a<0)$, for a given negative number $a$, there is a largest positive number $l(a)$, such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, l(a)]$.
Question: For what value of $a$ is $l(a)$ the largest? Find this largest $l(a)$ and prove your conclusion.

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Four. (This question is worth 20 points)
Let the function \( f(x) = a x^2 + 8 x + 3 \) (where \( a < 0 \)), for a given negative number \( a \), there is a largest positive number \( l(a) \), such that the inequality \( |f(x)| \leq 5 \) holds for the entire interval \([0, l(a)]\).
Question: For what value of \( a \) is \( l(a) \) the largest? Find this largest \( l(a) \) and prove your conclusion.
$$
\frac{\sqrt{5}+1}{2}
$$

Four, [Analysis and Solution] $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$. Therefore, $\max f(x)=3-\frac{16}{a}$. We discuss in two cases:
( i ) $3-\frac{16}{a}>5$, which means $-8-\frac{4}{a}$.

So, $l(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$,
$$
l(a)=\frac{-8-\sqrt{64-32 a}}{2 a}=\frac{2}{\sqrt{4-2 a-4}} \leqslant \frac{4}{\sqrt{20}-2}=\frac{\sqrt{5}+1}{2},
$$

with equality holding if and only if $a=-8$.
Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, the maximum value $\frac{\sqrt{5}+1}{2}$ is achieved if and only if $a=-8$.