Let $\lambda_1$ and $\lambda_2$ denote the eigenvalues of $\rho$. Then $\lambda_1 + \lambda_2 = \mathrm{tr}\rho = 1$ and $\lambda_1 \lambda_2 = \det \rho$. We can compute the determinant using the relationship between trace and determinant that underlies the Faddeev-LeVerrier algorithm

$$ \det\rho = \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right] = \frac{1}{2} - \frac{\mathrm{tr}\rho^2}{2}.\tag1 $$

We get

$$ \mathrm{tr}\rho^2 = \frac{1}{4}\mathrm{tr}(|a\rangle\langle a| + \langle a|b\rangle |a\rangle\langle b| + \langle b|a\rangle |b\rangle\langle a| + |b\rangle\langle b|) \\ = \frac{1}{2} + \frac{|\langle a|b\rangle|^2}{2} $$

and so

$$ \lambda_1 \lambda_2 = \det\rho = \frac{1}{4} - \frac{|\langle a|b\rangle|^2}{4} $$

from which we see that

$$ \lambda_i = \frac{1}{2} \pm \frac{|\langle a|b\rangle|}{2}. $$

We can prove $(1)$ as follows

$$ \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right] = \frac{1}{2}\left[(\lambda_1 + \lambda_2)^2 - (\lambda_1^2 + \lambda_2^2)\right] = \lambda_1\lambda_2 = \det\rho. $$

Let $\lambda_1$ and $\lambda_2$ denote the eigenvalues of $\rho$. Then $\lambda_1 + \lambda_2 = \mathrm{tr}\rho = 1$ and $\lambda_1 \lambda_2 = \det \rho$. We can compute the determinant using trace of $\rho$ and $\rho^2$

$$ \det\rho = \lambda_1\lambda_2 = \frac{1}{2}\left[(\lambda_1 + \lambda_2)^2 - (\lambda_1^2 + \lambda_2^2)\right] = \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right]. $$

(N.B. this useful relationship underlies the Faddeev-LeVerrier algorithm.) Now, calculate

$$ \mathrm{tr}\rho^2 = \frac{1}{4}\mathrm{tr}(|a\rangle\langle a| + \langle a|b\rangle |a\rangle\langle b| + \langle b|a\rangle |b\rangle\langle a| + |b\rangle\langle b|) \\ = \frac{1}{2} + \frac{|\langle a|b\rangle|^2}{2} $$

and so

$$ \lambda_1 \lambda_2 = \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right] = \frac{1}{4} - \frac{|\langle a|b\rangle|^2}{4} $$

from which we see that

$$ \lambda_i = \frac{1}{2} \pm \frac{|\langle a|b\rangle|}{2}. $$

Let $\lambda_1$ and $\lambda_2$ denote the eigenvalues of $\rho$. Then $\lambda_1 + \lambda_2 = \mathrm{tr}\rho = 1$ and $\lambda_1 \lambda_2 = \det \rho$. We can compute the determinant using the relationship between trace and determinant that underlies the Faddeev-LeVerrier algorithm

$$ \det\rho = \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right] = \frac{1}{2} - \frac{\mathrm{tr}\rho^2}{2}. $$

We get

$$ \mathrm{tr}\rho^2 = \frac{1}{4}\mathrm{tr}(|a\rangle\langle a| + \langle a|b\rangle |a\rangle\langle b| + \langle b|a\rangle |b\rangle\langle a| + |b\rangle\langle b|) \\ = \frac{1}{2} + \frac{|\langle a|b\rangle|^2}{2} $$

and so

$$ \lambda_1 \lambda_2 = \det\rho = \frac{1}{4} - \frac{|\langle a|b\rangle|^2}{4} $$

from which we see that

$$ \lambda_i = \frac{1}{2} \pm \frac{|\langle a|b\rangle|}{2}. $$

Let $\lambda_1$ and $\lambda_2$ denote the eigenvalues of $\rho$. Then $\lambda_1 + \lambda_2 = \mathrm{tr}\rho = 1$ and $\lambda_1 \lambda_2 = \det \rho$. We can compute the determinant using the relationship between trace and determinant that underlies the Faddeev-LeVerrier algorithm

$$ \det\rho = \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right] = \frac{1}{2} - \frac{\mathrm{tr}\rho^2}{2}.\tag1 $$

We get

$$ \mathrm{tr}\rho^2 = \frac{1}{4}\mathrm{tr}(|a\rangle\langle a| + \langle a|b\rangle |a\rangle\langle b| + \langle b|a\rangle |b\rangle\langle a| + |b\rangle\langle b|) \\ = \frac{1}{2} + \frac{|\langle a|b\rangle|^2}{2} $$

and so

$$ \lambda_1 \lambda_2 = \det\rho = \frac{1}{4} - \frac{|\langle a|b\rangle|^2}{4} $$

from which we see that

$$ \lambda_i = \frac{1}{2} \pm \frac{|\langle a|b\rangle|}{2}. $$

We can prove $(1)$ as follows

$$ \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right] = \frac{1}{2}\left[(\lambda_1 + \lambda_2)^2 - (\lambda_1^2 + \lambda_2^2)\right] = \lambda_1\lambda_2 = \det\rho. $$