1. način (djelomično korjenovanje unutar zagrade):
$$\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$$$\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$
$$\sqrt{3} + \sqrt{12} = \sqrt{3} + 2\sqrt{3} = 3\sqrt{3}$$$\sqrt{3} + \sqrt{12} = \sqrt{3} + 2\sqrt{3} = 3\sqrt{3}$
Kvadriramo dobiveni izraz:
$$(3\sqrt{3})^2 = 3^2 \cdot (\sqrt{3})^2 = 9 \cdot 3 = 27$$$(3\sqrt{3})^2 = 3^2 \cdot (\sqrt{3})^2 = 9 \cdot 3 = 27$

2. način (kvadrat binoma $(a+b)^2 = a^2 + 2ab + b^2$$(a+b)^2 = a^2 + 2ab + b^2$):
$$(\sqrt{3})^2 + 2\sqrt{3}\sqrt{12} + (\sqrt{12})^2 = 3 + 2\sqrt{36} + 12 = 3 + 2 \cdot 6 + 12 = 3 + 12 + 12 = 27$$$(\sqrt{3})^2 + 2\sqrt{3}\sqrt{12} + (\sqrt{12})^2 = 3 + 2\sqrt{36} + 12 = 3 + 2 \cdot 6 + 12 = 3 + 12 + 12 = 27$

Odgovor: **27**