= Solution
For the supplied system, the small-$\alpha$ <Taylor expansion> gives $\sqrt{1+2\alpha+5\alpha^2}=1+\alpha+O(\alpha^2)$, hence $\omega_-^2=\alpha g/l+O(\alpha^2g/l)$. Setting $\omega_-=(1/4)\sqrt{g/l}$ at this order yields
$$
\boxed{\alpha\simeq\frac1{16}.}
$$
This is an approximation in the squared <frequency>. The exact positive solution of $\lambda_-=1/16$ is $\alpha=(\sqrt{229}-13)/32\simeq0.06665$, showing the size of the neglected correction. The physical unequal-mass <double pendulum> has a different small-mass limit, so the source's requested tuning does not follow for that model.
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