x^2+4x+2=0
a=1, b=4, c=2
\Delta=b^2-4ac=16-4\cdot 1\cdot 2=8
\sqrt\Delta=\sqrt{8}=\sqrt{4\cdot 2}=2\sqrt2
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{-4-2\sqrt2}{2\cdot 1}=\frac{2(-2-\sqrt2)}{2}=-2-\sqrt2
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{-4+2\sqrt2}{2\cdot 1}=\frac{2(\sqrt2-2)}{2}=\sqrt2-2
Suma odwrotności liczb -2-\sqrt2 i \sqrt2-2
\frac{1}{-2-\sqrt2}+\frac{1}{\sqrt2-2}=\frac{1\cdot (-2+\sqrt2)}{(-2-\sqrt2)(-2+\sqrt2)}+\frac{1\cdot (\sqrt2+2)}{(\sqrt2-2)(\sqrt2+2)}=
=\frac{-2+\sqrt2}{(-2)^2-(\sqrt2)^2}+\frac{\sqrt2+2}{(\sqrt2)^2-2^2}=\frac{-2+\sqrt2}{4-2}+\frac{\sqrt2+2}{2-4}=
=\frac{-2+\sqrt2}{2}+\frac{\sqrt2+2}{-2}=\frac{-2+\sqrt2}{2}+\frac{-(\sqrt2+2)}{2}=
=\frac{-2+\sqrt2-\sqrt2-2}{2}=\frac{-4}{2}=-2 co należało wykazać