Question

Find the value of the square root \sqrt{6+2\sqrt{5}}


Collected in the board: Square Root

Steven Zheng posted 7 hours ago

Answer

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

a^2+5b^2=6
(1)

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

a^2 - 5b^2 =\pm4
(2)

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


Steven Zheng posted 7 hours ago

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