Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

simplify (√((3+2(√(2)))))

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

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

(3+2(√2) )^2 =((√2) )^2

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

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com