Question

if x= root3+root 2 by root3-root find x2+1byx2 and x4+1byx4​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

Now,

$\sf{x^2=\dfrac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})^2}}$

$\sf{\implies\,x^2=\dfrac{3+2+2\sqrt{6}}{3+2-2\sqrt{6}}}$

$\sf{\implies\,x^2=\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}}$

And,

$\sf{\implies\,x^4=\dfrac{(5+2\sqrt{6})^2}{(5-2\sqrt{6})^2}}$

$\sf{\implies\,x^4=\dfrac{25+24+20\sqrt{6}}{25+24-20\sqrt{6}}}$

$\sf{\implies\,x^4=\dfrac{49+20\sqrt{6}}{49-20\sqrt{6}}}$

Now,

$\sf{x^2+\dfrac{1}{x^2}=\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}+\dfrac{5-2\sqrt{6}}{5+2\sqrt{6}}}$

$\sf{\implies\,x^2+\dfrac{1}{x^2}=\dfrac{(5+2\sqrt{6})^2+(5-2\sqrt{6})^2}{(5-2\sqrt{6})(5+2\sqrt{6})}}$

$\sf{\implies\,x^2+\dfrac{1}{x^2}=\dfrac{2\{(5)^2+(2\sqrt{6})^2\}}{(5-2\sqrt{6})(5+2\sqrt{6})}}$

$\sf{\implies\,x^2+\dfrac{1}{x^2}=\dfrac{2(25+24)}{(5)^2-(2\sqrt{6})^2}}$

$\sf{\implies\,x^2+\dfrac{1}{x^2}=\dfrac{2\times49}{25-24}}$

$\sf{\implies\,x^2+\dfrac{1}{x^2}=\dfrac{2\times49}{1}}$

$\sf{\implies\,x^2+\dfrac{1}{x^2}=98}$