Question

if x=3+2√2 find the value of x square+1/xsquare
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Answer 1

Answer:

x = 3 + 2√2 ---( 1 )

1/x = 1/( 3 + 2√2 )

= ( 3 - 2√2 )/[ ( 3 + 2√2 )( 3 - 2√2 )]

= ( 3 - 2√2 )/ [ 3² - ( 2√2 )² ]

= ( 3 - 2√2 ) / ( 9 - 8 )

= 3 - 2√2 ----( 2 )

x + 1/x = 3 + 2√2 + 3 - 2√2

= 6 ----( 3 )

Now ,

do the Square of equation ( 3 ) , we get

( x + 1/x )² = 6²

x² + 1/x² + 2 = 36

x² + 1/x² = 36 - 2

= 34

I hope this helps you.

: )

Answer 2

❏ SolutioN :

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\blacksquare\:\:\footnotesize{\underline{\underline{Given}}}$

$\footnotesize{X=3+2\sqrt{2}}$

$\blacksquare\:\:\footnotesize{\underline{\underline{To\: Find}}}$

$\footnotesize{(x^2+\dfrac{1}{x^2})=?}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\longrightarrow\footnotesize{X=3+2\sqrt{2}}$

$\longrightarrow\footnotesize{\dfrac{1}{X}=\dfrac{1}{3+2\sqrt{2}}}$

$\longrightarrow\footnotesize{\dfrac{1}{X}=\dfrac{1(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}}$

$\longrightarrow\footnotesize{\dfrac{1}{X}=\dfrac{1(3-2\sqrt{2})}{3^2-(2\sqrt{2})^2}}$

$\longrightarrow\footnotesize{\dfrac{1}{X}=\dfrac{1(3-2\sqrt{2})}{9-8}}$

$\longrightarrow\footnotesize{\dfrac{1}{X}=(3-2\sqrt{2})}$

$\therefore\:\footnotesize{X^2+\dfrac{1}{X^2}}$

$\implies\footnotesize{(X+\dfrac{1}{X})^2-2\times X\times\dfrac{1}{X}}$

$\implies\footnotesize{(X+\dfrac{1}{X})^2-2}$

$\footnotesize{putting\:values\:X\:and\:\dfrac{1}{X}\:,\,we\: get }$

$\implies\footnotesize{[(3+2\sqrt{2})+(3-2\sqrt{2})]^2-2}$

$\implies\footnotesize{[3+\cancel{2\sqrt{2}}+3-\cancel{2\sqrt{2}}]^2-2}$

$\implies\footnotesize{(3+3)^2-2}$

$\implies\footnotesize{(6)^2-2}$

$\implies\footnotesize{36-2}$

$\implies\footnotesize{34}$

$\therefore\:\:\:\boxed{\footnotesize{(X^2+\dfrac{1}{X^2})=34}}$ if $\footnotesize{X=3+2\sqrt{2}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\put(1,1.2){\line(1,0){6.5}}\end{picture}$