Question

Find the value of the square root \sqrt{9+4\sqrt{2}}


Collected in the board: Square Root

Steven Zheng posted 5 hours ago

Answer

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

a^2+2b^2=9
(1)

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

a^2 - 2b^2 =\pm7
(2)

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


Steven Zheng posted 5 hours ago

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