Question
Find the value of the square root \sqrt{9+4\sqrt{2}}
Find the value of the square root \sqrt{9+4\sqrt{2}}
Let
9+4\sqrt{2} = (a+b\sqrt{2})^2
9-4\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 = 18
Expand and simplify the equation
2a^2+2\cdot 2b^2=18
Multiplying the two equations gives
(9+4\sqrt{2})(9-4\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 = 9^2 - 4^2\cdot 2=49
Taking the square root
2 systems of equations are obtained from (1) and (2)
Case 1
\begin{cases} a^2+2b^2&=9 \\ a^2 - 2b^2 &=7 \end{cases}
Solving the system of equaitons gives
a = \sqrt{\dfrac{9+7}{2}} =\sqrt{8} = 2\sqrt{2}
b = \sqrt{\dfrac{9-7}{2\cdot 2}}=\sqrt{\dfrac{1}{2}}
Case 2
\begin{cases} a^2+2b^2&=9\\ a^2 - 2b^2 &= -7 \end{cases}
Solving the system of equaitons gives
a = 1, b=2
Both results give the same root. Therefore,
The value of the square root of
\sqrt{9+4\sqrt{2}}=\sqrt{(1+2\sqrt{2})^2}=1+2\sqrt{2}
\sqrt{9-4\sqrt{2}}=\sqrt{(1-2\sqrt{2})^2}=2\sqrt{2}-1