Question

Find the value of $x^2+\frac{1}{x^2}$, when $x=1-\sqrt{2}$,

• A. 6
• B. 16
• C. 24
• D. none of the above

Answer 1

The correct option is A 6

It is given that $x=1-\sqrt{2}$.

Let us first find $x^2$ as follows:

$x^2={(1-\sqrt{2})}^2={\left(1\right)}^2+{\left(\sqrt{2}\right)}^2-(2\times 1\times \left(\sqrt{2}\right))\{\because {(a-b)}^2=({a^2+b}^2-2ab)\}=1+2-2\sqrt{2}=3-2\sqrt{2}$

Therefore, $x^2=3-2\sqrt{2}$

Now consider $x^2+\frac{1}{x^2}$ as follows:

$x^2+\frac{1}{x^2}=3-2\sqrt{2}+\frac{1}{3-2\sqrt{2}}=\frac{{(3-2\sqrt{2})}^2+1}{3-2\sqrt{2}}=\frac{[{\left(3\right)}^2+{\left(2\sqrt{2}\right)}^2-(2\times 3\times \left(2\sqrt{2}\right))]+1}{3-2\sqrt{2}}\{\because {(a-b)}^2=({a^2+b}^2-2ab)\}=\frac{[9+(4\times 2)-12\sqrt{2}]+1}{3-2\sqrt{2}}$

$=\frac{9+8-12\sqrt{2}+1}{3-2\sqrt{2}}=\frac{18-12\sqrt{2}}{3-2\sqrt{2}}=\frac{18-12\sqrt{2}}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}\left\{Rationalizedenominator\right\}=\frac{(18-12\sqrt{2})(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}$

$=\frac{(18\times 3)+(18\times 2\sqrt{2})+(-12\sqrt{2}\times 3)+(-12\sqrt{2}\times 2\sqrt{2})}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}\{\because (a-b)(a+b)=a^2-b^2\}=\frac{54+36\sqrt{2}-36\sqrt{2}-48}{9-8}=\frac{6}{1}=6$

Hence, $x^2+\frac{1}{x^2}=6$ when $x=1-\sqrt{2}$.