Question

If $\bullet a=3+\sqrt{8}$

Then find

$\bullet \bigg(a{}^{2}+\dfrac{1}{a{}^{2}}\bigg)$​

Answer 1

To Find :-

• Value of $\sf \bigg(a{}^{2}+\dfrac{1}{a{}^{2}}\bigg) =?$

Solution :-

Given,

• $\sf a=3+\sqrt{8}$

$\sf Then$

$\sf\bigg(\dfrac{1}{a}\bigg)=\dfrac{1}{3+\sqrt{8}}$

$\sf:\implies \dfrac{1}{3+\sqrt{8}}\times \dfrac{(3-\sqrt{8})}{(3-\sqrt{8})}$

$\sf:\implies \dfrac{1}{a}=\dfrac{3-\sqrt{8}}{(3){}^{2}-(\sqrt{8}){}^{2}}$

$\sf:\implies \dfrac{1}{a}=\dfrac{3-\sqrt{8}}{9-8}$

$\sf:\implies \dfrac{1}{a}= 3-\sqrt{8}$

$\sf Now$

$\sf \bigg(a{}^{2}+\dfrac{1}{a{}^{2}}\bigg)$

$\sf:\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}+2$

$\sf:\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}=( 3+\cancel{\sqrt{8}}+3-\cancel{\sqrt{8}}){}^{2}$

$\sf:\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}= (6){}^{2}=36$

$\sf:\implies 36=a{}^{2}+\dfrac{1}{a{}^{2}}+2$

$\sf:\implies 36-2=a{}^{2}+\dfrac{1}{a{}^{2}}$

$\sf:\implies 34=a{}^{2}+\dfrac{1}{a{}^{2}}$

$\sf:\implies a{}^{2}+\dfrac{1}{a{}^{2}}= 34$

$\boxed {\sf {\purple {Required \ value \ is\ 34}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

$\sf \qquad \green{\bullet} \: Value \ of \ a \ is \ 3+\sqrt{8}$

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:FinD:-}}}}}$

$\sf \qquad \green{\bullet}\: The \ value \ of \ \bigg\lgroup a^2 + \dfrac{1}{a^2}\bigg\rgroup$

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } AnsweR:-}}}}}$

$\sf Given \ that :-$

$\tt:\implies a = 3 + \sqrt{8} \\\\\tt:\implies \dfrac{1}{a}= \dfrac{1}{3+\sqrt{8}} \\\\\tt:\implies\dfrac{1}{a}=\bigg\lgroup \dfrac{(3-\sqrt{8})}{(3+\sqrt{8})(3-\sqrt{8})}\bigg\rgroup \\\\\tt:\implies\dfrac{1}{a}=\bigg\lgroup \dfrac{3-\sqrt{8}}{(3)^2-(\sqrt8)^2}\bigg\rgroup \\\\\underline{\boxed{\red{\tt\longmapsto \dfrac{1}{a}=3-\sqrt{8} }}}$

$\rule{200}2$

$\underline{\pink{\sf On \ adding \ them \ we \ have :- }}$

$\tt:\implies a + \dfrac{1}{a}= 3 + \sqrt8 + 3 -\sqrt8 \\\\\tt:\implies a + \dfrac{1}{a} = 6 \\\\\tt:\implies \bigg\lgroup a + \dfrac{1}{a}\bigg\rgroup ^2 = 6^2\\\\\tt:\implies \bigg\lgroup a^2 + \dfrac{1}{a^2}+2 \times a \times \dfrac{1}{ a } \bigg\rgroup = 36 \\\\\tt:\implies </p><p> \bigg\lgroup a^2 + \dfrac{1}{a^2}\bigg\rgroup + 2 = 36 \\\\\tt:\implies \bigg\lgroup a^2 + \dfrac{1}{a^2}\bigg\rgroup = 36 - 2 \\\\\underline{\boxed{\red{\tt\longmapsto \bigg\lgroup a^2 + \dfrac{1}{a^2}\bigg\rgroup=34 }}}$

$\boxed{\green{\bf \pink{\dag}\: Hence \ the \ required \ answer \ is \ 34 . }}$