Answer
$\frac{w+\sqrt{5{{w}^{2}}}}{2}$
Work Step by Step
$\frac{w}{l}=\frac{l}{w+l}$
Solving the equation for $l$,
Multiply $l\left( w+l \right)$ on both sides of the equation,
$\begin{align}
& \frac{w}{l}=\frac{l}{w+l} \\
& \frac{w}{l}\times l\left( w+l \right)=\frac{l}{w+l}\times l\left( w+l \right) \\
& w\left( w+l \right)={{l}^{2}} \\
& {{w}^{2}}+wl={{l}^{2}}
\end{align}$
Rearrange the equation,
$0={{l}^{2}}-lw-{{w}^{2}}$
Using the quadratic formula,
$\begin{align}
& l=\frac{-\left( -w \right)\pm \sqrt{{{\left( -w \right)}^{2}}-4\times 1\left( -{{w}^{2}} \right)}}{2\times 1} \\
& =\frac{w\pm \sqrt{{{w}^{2}}+4{{w}^{2}}}}{2} \\
& =\frac{w\pm \sqrt{5{{w}^{2}}}}{2}
\end{align}$
Length cannot be negative, so we only consider the positive roots.
$l=\frac{w+\sqrt{5{{w}^{2}}}}{2}$
Thus, the value is $l=\frac{w+\sqrt{5{{w}^{2}}}}{2}$.
