Question

If (x=2+root3 ),find the value of (x square + 1/x square)square

Answer 1

$\huge\underline{\underline{\red{\tt{ANSWER:-}}}}$

$\sf\underline{\purple{Given:-}}$

$\sf x=2+\sqrt{3}$

We have to find the value of

$\sf x{}^{2}+\dfrac{1}{x{}^{2}}$

Now :-

• Find the value of 1/x

$\sf\sf\:\:\:\: \dfrac{1}{x}=\dfrac{1}{2+\sqrt{3}}\\ \\ \sf\implies \dfrac{1}{x}=\dfrac{1}{2+\sqrt{3}}\times \dfrac{(2-\sqrt{3})}{(2-\sqrt{3})}\\ \\ \sf\implies \dfrac{1}{x}=\dfrac{2-\sqrt{3}}{(2){}^{2}-(\sqrt{3}){}^{2}}\\ \\ \sf\implies \dfrac{1}{x}=\dfrac{2-\sqrt{3}}{4-3}\\ \\ \sf\implies \dfrac{1}{x}=\dfrac{2-\sqrt{3}}{1}=2-\sqrt{3}$

• The value of 1/x=2-√3

Now the value of $\sf x+\dfrac{1}{x}$

$\sf\implies x+\dfrac{1}{x}\\ \\ \sf\implies x+ \dfrac{1}{x}=2+\cancel{\sqrt{3}}+2-\cancel{\sqrt{3}}\\ \\ \sf\implies x+\dfrac{1}{x}= 4$

Now the value of $\sf x{}^{2}+\dfrac{1}{x{}^{2}}$

$\sf \implies \bigg ( x+\dfrac{1}{x} \bigg){}^{2}=x{}^{2}+\dfrac {1}{x{}^{2}}+2\\ \\ \sf\implies (4){}^{2}=x{}^{2}+\dfrac{1}{x{}^{2}}+2\\ \\ \sf\implies 16-2=x{}^{2}+\dfrac{1}{x{}^{2}}\\ \\ \sf\implies x{}^{2}+\dfrac{1}{x{}^{2}}=14$

$\boxed{\mathfrak{\red{ x{}^{2}+\dfrac{1}{x{}^{2}}=14}}}$

Answer 2

Answer:

14

Step-by-step explanation:

[x+1/x]²=x²+1/x²+2×x×1/x