Question

√(11 - 6 √(2))
(not decimal form)

Answer 1

`\huge\boxed{√(11-6\sqrt2)=3-\sqrt2}`

Explanation:

`√(11-6\sqrt2)=(*)\\\\11=9+2=3^2+(\sqrt2)^2\\6\sqrt2=(2)(3)(\sqrt2)\\\\(*)=√(3^2-(2)(3)(\sqrt2)+(\sqrt2)^2)\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=√((3-\sqrt2)^2)\\\\\text{use}\ √(a^2)=|a|\\\\=|3-\sqrt2|=3-\sqrt2\\\\^{\text{because}\ 3>\sqrt2\to3-\sqrt2>0}`