Question

Find the value of the square roo \sqrt{21+4\sqrt{5}}


Collected in the board: Square Root

Steven Zheng posted 7 hours ago

Answer

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

a^2+5b^2=21
(1)

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

a^2 - 5b^2 =\pm19
(2)

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


Steven Zheng posted 7 hours ago

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