Question
Find the value of the square roo \sqrt{21+4\sqrt{5}}
Find the value of the square roo \sqrt{21+4\sqrt{5}}
Let
21+4\sqrt{5} = (a+b\sqrt{5})^2
21-4\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 = 42
Expand and simplify the equation
2a^2+2\cdot 5b^2=42
Multiplying the two equations gives
(21+4\sqrt{5})(21-4\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 = 21^2 - 4^2\cdot 5=361
Taking the square root
2 systems of equations are obtained from (1) and (2)
Case 1
\begin{cases} a^2+5b^2&=21 \\ a^2 - 5b^2 &=19 \end{cases}
Solving the system of equaitons gives
a = \sqrt{\dfrac{21+19}{2}} =\sqrt{20} = 2\sqrt{5}
b = \sqrt{\dfrac{21-19}{2\cdot 5}}=\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{5}}
Case 2
\begin{cases} a^2+5b^2&=21\\ a^2 - 5b^2 &= -19 \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{21+4\sqrt{5}}=\sqrt{(1+2\sqrt{5})^2}=1+2\sqrt{5}
\sqrt{21-4\sqrt{5}}=\sqrt{(1-2\sqrt{5})^2}=2\sqrt{5}-1