Let

3+2\sqrt{2} = (a+b\sqrt{2})^2

3-2\sqrt{2} = (a-b\sqrt{2})^2

in which a and b are positive integers

Addition of the two equations gives

(a+b\sqrt{2})^2+(a-b\sqrt{2})^2 = 6

Expand and simplify the equation

2a^2+2\cdot 2b^2=6

Multiplying the two equations gives

(3+2\sqrt{2})(3-2\sqrt{2}) = (a+b\sqrt{2})^2 (a-b\sqrt{2})^2

Simplify the equation using difference of square formula

(a^2 - b^2\cdot 2)^2 = 3^2 - 2^2\cdot 2=1

Taking the square root

2 systems of equations are obtained from (1) and (2)

Case 1

\begin{cases} a^2+2b^2&=3 \\ a^2 - 2b^2 &=1 \end{cases}

Solving the system of equaitons gives

a = \sqrt{\dfrac{3+1}{2}} =\sqrt{2}

b = \sqrt{\dfrac{3-1}{2\cdot 2}}=\sqrt{\dfrac{1}{2}}

Case 2

\begin{cases} a^2+2b^2&=3\\ a^2 - 2b^2 &= -1 \end{cases}

Solving the system of equaitons gives

a = 1, b=1

Both results give the same root. Therefore,

The value of the square root of

\sqrt{3+2\sqrt{2}}=\sqrt{(1+\sqrt{2})^2}=1+\sqrt{2}

\sqrt{3-2\sqrt{2}}=\sqrt{(1-\sqrt{2})^2}=\sqrt{2}-1