б) x² + (v11 + v2)x + V22 = 0

$$x^2 + (\sqrt{11} + \sqrt{2})x + \sqrt{22} = 0$$

Найдем дискриминант:

$$D = (\sqrt{11} + \sqrt{2})^2 - 4 \cdot 1 \cdot \sqrt{22} = 11 + 2\sqrt{22} + 2 - 4\sqrt{22} = 13 - 2\sqrt{22}$$

$$x_{1,2} = \frac{-(\sqrt{11} + \sqrt{2}) \pm \sqrt{13 - 2\sqrt{22}}}{2}$$

Обратим внимание, что $$13 - 2\sqrt{22} = (\sqrt{11} - \sqrt{2})^2$$

$$x_{1,2} = \frac{-(\sqrt{11} + \sqrt{2}) \pm (\sqrt{11} - \sqrt{2})}{2}$$

$$x_1 = \frac{-\sqrt{11} - \sqrt{2} + \sqrt{11} - \sqrt{2}}{2} = \frac{-2\sqrt{2}}{2} = -\sqrt{2}$$

$$x_2 = \frac{-\sqrt{11} - \sqrt{2} - \sqrt{11} + \sqrt{2}}{2} = \frac{-2\sqrt{11}}{2} = -\sqrt{11}$$

Ответ: $$x_1 = -\sqrt{2}$$, $$x_2 = -\sqrt{11}$$