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

(√(2+(√3))) + (√(2−(√3))) = ?

$$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:+\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:? \\ $$

Answered by 123456 last updated on 11/Feb/15

x=(√(2+(√3)))+(√(2−(√3)))  x^2 =2+(√3)+2(√(2+(√3)))(√(2−(√3)))+2−(√3)  x^2 =4+2(√((2+(√3))(2−(√3))))  x^2 =4+2(√(4−3))=4+2=6  x=±(√6)  3<2+(√3)<4⇔1<(√3)<2⇔−2<−(√3)<−1⇔0<2−(√3)<1  2+(√3)>0  2−(√3)>0  x=(√6)

$${x}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}+\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} =\mathrm{4}+\mathrm{2}\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)} \\ $$$${x}^{\mathrm{2}} =\mathrm{4}+\mathrm{2}\sqrt{\mathrm{4}−\mathrm{3}}=\mathrm{4}+\mathrm{2}=\mathrm{6} \\ $$$${x}=\pm\sqrt{\mathrm{6}} \\ $$$$\mathrm{3}<\mathrm{2}+\sqrt{\mathrm{3}}<\mathrm{4}\Leftrightarrow\mathrm{1}<\sqrt{\mathrm{3}}<\mathrm{2}\Leftrightarrow−\mathrm{2}<−\sqrt{\mathrm{3}}<−\mathrm{1}\Leftrightarrow\mathrm{0}<\mathrm{2}−\sqrt{\mathrm{3}}<\mathrm{1} \\ $$$$\mathrm{2}+\sqrt{\mathrm{3}}>\mathrm{0} \\ $$$$\mathrm{2}−\sqrt{\mathrm{3}}>\mathrm{0} \\ $$$${x}=\sqrt{\mathrm{6}} \\ $$

Commented by 123456 last updated on 11/Feb/15

y=(√(2+(√3)))−(√(2−(√3)))  y^2 =2+(√3)−2(√(2+(√3)))(√(2−(√3)))+2−(√3)  y^2 =4−2(√((2+(√3))(2−(√3))))  y^2 =4−2(√(4−3))=4−2=2  y=±(√2)  {−(√6),−(√2),+(√2),+(√6)}  (√3)>−(√3)>−2  2+(√3)>2−(√3)>0  (√(2+(√3)))>(√(2−(√3)))>0

$${y}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}−\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}+\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}−\mathrm{2}\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}−\mathrm{2}\sqrt{\mathrm{4}−\mathrm{3}}=\mathrm{4}−\mathrm{2}=\mathrm{2} \\ $$$${y}=\pm\sqrt{\mathrm{2}} \\ $$$$\left\{−\sqrt{\mathrm{6}},−\sqrt{\mathrm{2}},+\sqrt{\mathrm{2}},+\sqrt{\mathrm{6}}\right\} \\ $$$$\sqrt{\mathrm{3}}>−\sqrt{\mathrm{3}}>−\mathrm{2} \\ $$$$\mathrm{2}+\sqrt{\mathrm{3}}>\mathrm{2}−\sqrt{\mathrm{3}}>\mathrm{0} \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}>\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}>\mathrm{0} \\ $$

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