Question
Find the value of the square root \sqrt{6+2\sqrt{5}}
Find the value of the square root \sqrt{6+2\sqrt{5}}
Let
6+2\sqrt{5} = (a+b\sqrt{5})^2
6-2\sqrt{5} = (a-b\sqrt{5})^2
in which a and b are positive integers
Addition of the two equations gives
(a+b\sqrt{5})^2+(a-b\sqrt{5})^2 = 12
Expand and simplify the equation
2a^2+2\cdot 5b^2=12
Multiplying the two equations gives
(6+2\sqrt{5})(6-2\sqrt{5}) = (a+b\sqrt{5})^2 (a-b\sqrt{5})^2
Simplify the equation using difference of square formula
(a^2 - b^2\cdot 5)^2 = 6^2 - 2^2\cdot 5=16
Taking the square root
2 systems of equations are obtained from (1) and (2)
Case 1
\begin{cases} a^2+5b^2&=6 \\ a^2 - 5b^2 &=4 \end{cases}
Solving the system of equaitons gives
a = \sqrt{\dfrac{6+4}{2}} =\sqrt{5} = \sqrt{5}
b = \sqrt{\dfrac{6-4}{2\cdot 5}}=\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{5}}
Case 2
\begin{cases} a^2+5b^2&=6\\ a^2 - 5b^2 &= -4 \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{6+2\sqrt{5}}=\sqrt{(1+\sqrt{5})^2}=1+\sqrt{5}
\sqrt{6-2\sqrt{5}}=\sqrt{(1-\sqrt{5})^2}=\sqrt{5}-1