Let

17+12\sqrt{2} = (a+b\sqrt{2})^2

17-12\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 = 34

Expand and simplify the equation

2a^2+2\cdot 2b^2=34

Multiplying the two equations gives

(17+12\sqrt{2})(17-12\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 = 17^2 - 12^2\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&=17 \\ a^2 - 2b^2 &=1 \end{cases}

Solving the system of equaitons gives

a = \sqrt{\dfrac{17+1}{2}} =\sqrt{9} = 3

b = \sqrt{\dfrac{17-1}{2\cdot 2}}=\sqrt{4}= 2

Case 2

\begin{cases} a^2+2b^2&=17\\ a^2 - 2b^2 &= -1 \end{cases}

Solving the system of equaitons gives

a = 2\sqrt{2}, b=3\sqrt{\dfrac{1}{2} }

Both results give the same root. Therefore,

The value of the square root of

\sqrt{17+12\sqrt{2}}=\sqrt{(3+2\sqrt{2})^2}=3+2\sqrt{2}

\sqrt{17-12\sqrt{2}}=\sqrt{(3-2\sqrt{2})^2}=3-2\sqrt{2}