Question

If a = 3 + root 8 find the value of a square + 1 upon a square

Answer 1

Answer:

hope you understood thanks.. to find:a2 plus 1/a2

Answer 2

$\huge\mathbb{SOLUTION:-}$

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

Now find the Value of

$\sf\bigg(\dfrac{1}{a}\bigg)=\dfrac{1}{3+\sqrt{8}}\\ \\ \sf\implies \dfrac{1}{3+\sqrt{8}}\times \dfrac{(3-\sqrt{8})}{(3-\sqrt{8})}\\ \\ \sf\implies \dfrac{1}{a}=\dfrac{3-\sqrt{8}}{(3){}^{2}-(\sqrt{8}){}^{2}}\\ \\ \sf\implies \dfrac{1}{a}=\dfrac{3-\sqrt{8}}{9-8}\\ \\ \sf\implies \dfrac{1}{a}= 3-\sqrt{8}$

Now find the value of

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

$\boxed{\sf{\red{(a+b){}^{2}= a{}^{2}+2ab+b{}^{2}}}}$

$\sf\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}+2\\ \\ \sf\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}=( 3+\cancel{\sqrt{8}}+3-\cancel{\sqrt{8}}){}^{2}\\ \\ \sf\implies \bigg(a+\dfrac{1}{a}\bigg){}^{2}= (6){}^{2}=9\\ \\ \sf\implies 9=a{}^{2}+\dfrac{1}{a{}^{2}}+2\\ \\ \sf\implies 9-2=a{}^{2}+\dfrac{1}{a{}^{2}}\\ \\ \sf\implies 7=a{}^{2}+\dfrac{1}{a{}^{2}}$

$\boxed{\sf{a{}^{2}+\dfrac{1}{a{}^{2}}=7}}$