2-3-2-3-

2-3-2-3-

Question Number 619 by Cheenz last updated on 11/Feb/15

\sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}} = ?

Answered by 123456 last updated on 11/Feb/15

x = \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}}

x^{2} = 2 + \sqrt{3} + 2 \sqrt{2 + \sqrt{3}} \sqrt{2 - \sqrt{3}} + 2 - \sqrt{3}

x^{2} = 4 + 2 \sqrt{(2 + \sqrt{3}) (2 - \sqrt{3})}

x^{2} = 4 + 2 \sqrt{4 - 3} = 4 + 2 = 6

x = \pm \sqrt{6}

3 < 2 + \sqrt{3} < 4 \Leftrightarrow 1 < \sqrt{3} < 2 \Leftrightarrow - 2 < - \sqrt{3} < - 1 \Leftrightarrow 0 < 2 - \sqrt{3} < 1

2 + \sqrt{3} > 0

2 - \sqrt{3} > 0

x = \sqrt{6}

Commented by 123456 last updated on 11/Feb/15

y = \sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}

y^{2} = 2 + \sqrt{3} - 2 \sqrt{2 + \sqrt{3}} \sqrt{2 - \sqrt{3}} + 2 - \sqrt{3}

y^{2} = 4 - 2 \sqrt{(2 + \sqrt{3}) (2 - \sqrt{3})}

y^{2} = 4 - 2 \sqrt{4 - 3} = 4 - 2 = 2

y = \pm \sqrt{2}

{- \sqrt{6}, - \sqrt{2}, + \sqrt{2}, + \sqrt{6}}

\sqrt{3} > - \sqrt{3} > - 2

2 + \sqrt{3} > 2 - \sqrt{3} > 0

\sqrt{2 + \sqrt{3}} > \sqrt{2 - \sqrt{3}} > 0

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