Question

If $x=3-2\sqrt{2}$, find the value of $x^4+\frac{1}{x^4}$.

Answer 1

It is given that $x=3-2\sqrt{2}$, and we have to find find $x^4$, so we first find $x^2$ that is ${(3-2\sqrt{2})}^2$ as follows:

${(3-2\sqrt{2})}^2=\{{\left(3\right)}^2+{\left(2\sqrt{2}\right)}^2-(2\times 3\times 2\sqrt{2})\}\{\because {(a-b)}^2=({a^2+b}^2-2ab)\}=\{9+8-\left(12\sqrt{2}\right)\}=17-12\sqrt{2}$

Therefore, $x^2=17-12\sqrt{2}$

Now, let us find $x^4=(x^2{)}^2$ that is:

${(17-12\sqrt{2})}^2=\{{\left(17\right)}^2+{\left(12\sqrt{2}\right)}^2-(2\times 17\times 12\sqrt{2})\}\{\because {(a-b)}^2=({a^2+b}^2-2ab)\}=\{289+288-\left(408\sqrt{2}\right)\}=577-408\sqrt{2}$

Thus, $x^4=577-408\sqrt{2}$

Now we will find the value of $x^4+\frac{1}{x^4}$ as shown below:

$x^4+\frac{1}{x^4}=577-408\sqrt{2}+\left(\frac{1}{577-408\sqrt{2}}\right)=577-408\sqrt{2}+(\frac{1}{577-408\sqrt{2}}\times \frac{577+408\sqrt{2}}{577+408\sqrt{2}})\left\{Rationalizedenominator\right\}=577-408\sqrt{2}+\left(\frac{577+408\sqrt{2}}{(577-408\sqrt{2})(577+408\sqrt{2})}\right)=577-408\sqrt{2}+\left(\frac{577+408\sqrt{2}}{{\left(577\right)}^2-{\left(408\sqrt{2}\right)}^2}\right)\{\because (a-b)(a+b)=a^2-b^2\}$

$=577-408\sqrt{2}+\left(\frac{577+408\sqrt{2}}{332929-332928}\right)=577-408\sqrt{2}+\left(\frac{577+408\sqrt{2}}{1}\right)=577-408\sqrt{2}+577+408\sqrt{2}=577+577=1154$

Hence, $x^4+\frac{1}{x^4}=1154$