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You got to the quadratic equation $$\lambda^2 - (\rho - n)\lambda + \frac{c^\ast f''(k^\ast)}{\varepsilon}$$ The discriminant is given by: $$\Delta = (\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{\varepsilon}$$ So the two roots are: $$\lambda_1, \lambda_2 = \frac{(\rho - n) \pm \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}}{2}$$ As $f''(k) < 0$ ...