Question

simplify:
$\sqrt{3 + 2 \sqrt{2} }$
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Answer 1

Answer:

Solution:

$\sf{Let \ x =\sqrt{3+2\sqrt2}} \\ \\ \sf{On \ squaring \ both \ sides} \\ \\ \sf{x^{2}=3+2\sqrt2} \\ \\ \sf{\therefore{x^{2}-3=2\sqrt2}} \\ \\ \sf{On \ squaring \ both \ sides} \\ \\ \sf{x^{2}-3x+9=8} \\ \\ \sf{\therefore{x^{4}-6x^{2}+1=0}} \\ \\ \sf{Substitute \ x^{2}=a} \\ \\ \sf{a^{2}-6a+1=0} \\ \\ \sf{By \ formula \ method} \\ \\ \sf{a=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}} \\ \\ \sf{\therefore{a=\dfrac{6\pm\sqrt{6^{2}-4(1)(1)}}{2}}} \\ \\ \sf{\therefore{a=\dfrac{6\pm4\sqrt2}{2}}}$

$\sf{\therefore{a=3\pm2\sqrt2}} \\ \\ \sf{But \ a \ can't \ be \ 3-2\sqrt2 \ since \ their} \\ \sf{no \ negative \ sign \ inside \ the \ root.} \\ \\ \sf{a=3+2\sqrt2} \\ \\ \sf{\therefore{a=1+2+2\sqrt2}} \\ \\ \boxed{\sf{a^{2}+b^{2}+2ab=(a+b)^{2}}} \\ \\ \sf{\therefore{a=(1+\sqrt2)^{2}}} \\ \\ \sf{\therefore{x^{2}=(1+\sqrt2)^{2}}} \\ \\ \sf{x=1+\sqrt2}$

$\sf{Hence, \ the \ simplified \ form \ is \ 1+\sqrt2.}$