simplify-3-2-2-

Question Number 33575 by MURALI last updated on 19/Apr/18

$$\mathrm{simplify}\:\sqrt{\left(\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.} \\ $$

Answered by Rasheed.Sindhi last updated on 19/Apr/18

Let (√((3+2(√(2))))) =a+b(√2) where a,b ∈Z ( (√((3+2(√(2))))) )^2 =(a+b(√2) )^2 a^2 +2b^2 +2ab(√2)=3+2(√2) a^2 +2b^2 =3 ∧ ab=1 ab=1⇒b=1/a a^2 +2b^2 =3⇒a^2 +(2/a^2 )=3 a^4 −3a^2 +2=0 (a^2 −1)(a^2 −2)=0 a=±1 , a=±(√2) ∉Z Hence a=±1 b=1/(±1)=±1 ∴ (√((3+2(√(2)))))=1+(√2) , −1−(√2)

$$\mathrm{Let}\:\sqrt{\left(\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.}\:\:=\mathrm{a}+\mathrm{b}\sqrt{\mathrm{2}} \\ $$$$\mathrm{where}\:\mathrm{a},\mathrm{b}\:\in\mathbb{Z} \\ $$$$\:\:\:\left(\:\sqrt{\left(\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.}\:\right)^{\mathrm{2}} =\left(\mathrm{a}+\mathrm{b}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{2b}^{\mathrm{2}} +\mathrm{2ab}\sqrt{\mathrm{2}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{2b}^{\mathrm{2}} =\mathrm{3}\:\wedge\:\:\mathrm{ab}=\mathrm{1} \\ $$$$\mathrm{ab}=\mathrm{1}\Rightarrow\mathrm{b}=\mathrm{1}/\mathrm{a} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{2b}^{\mathrm{2}} =\mathrm{3}\Rightarrow\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} }=\mathrm{3} \\ $$$$\mathrm{a}^{\mathrm{4}} −\mathrm{3a}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{a}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{a}=\pm\mathrm{1}\:,\:\mathrm{a}=\pm\sqrt{\mathrm{2}}\:\notin\mathbb{Z} \\ $$$$\mathrm{Hence}\:\mathrm{a}=\pm\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}=\mathrm{1}/\left(\pm\mathrm{1}\right)=\pm\mathrm{1} \\ $$$$\therefore\:\:\:\sqrt{\left(\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.}=\mathrm{1}+\sqrt{\mathrm{2}}\:\:,\:\:−\mathrm{1}−\sqrt{\mathrm{2}} \\ $$

Answered by MURALI last updated on 19/Apr/18

$$\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\right)^{\mathrm{2}} \:=\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 21/Apr/18

Perhaps you mean (3+2(√2) )^2 =(1+(√2) )^2

$$\mathrm{Perhaps}\:\mathrm{you}\:\mathrm{mean} \\ $$$$\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\right)^{\mathrm{2}} \:=\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$$ \\ $$

Answered by MURALI last updated on 19/Apr/18

(3+ 2(√2) )=((√2) )^2 +2(√2) ×1+(1)^2 (3+ 2(√2) )=((√2) +1)^2 ∴ (√(3+2(√(2))))) = (√2) +1

$$\left(\mathrm{3}+\:\mathrm{2}\sqrt{\mathrm{2}}\:\right)=\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\:\:×\mathrm{1}+\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{3}+\:\mathrm{2}\sqrt{\mathrm{2}}\:\right)=\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\therefore\:\:\sqrt{\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}}\:\:=\:\:\sqrt{\mathrm{2}}\:+\mathrm{1} \\ $$$$ \\ $$

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