Question 958167
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For clarity, let x=l (to prevent confusion with one)
{{{w/x=x/(w+x)}}} Put value in for w:
{{{1cm/x=x/(1cm+x)}}} Multiply by x
{{{1cm=(x^2)/(1cm+x)}}} Multiply by 1cm+x
{{{x+1=x^2}}} Subtract (x+1)from each side.
{{{0=x^2-x-1}}}
*[invoke quadratic "x", 1, -1, -1 ]
ANSWER: x=1.62 or -0.62 (Remember x is l)