Question

If $x=(7+4\sqrt{3})$, then the value of $x^2+\frac{1}{x^2}$ is:

• A. $193$
• B. $194$
• C. $195$
• D. $196$

Answer 1

The correct option is B $194$

Consider $x=7+4\sqrt{3}$ and calculate $x^2$ as follows:

$x^2={(7+4\sqrt{3})}^2=49+48+56\sqrt{3}=97+56\sqrt{3}$ .

Now $\frac{1}{x^2}$ can be determined as follows:

$\frac{1}{x^2}=\frac{1}{97+56\sqrt{3}}\times \frac{97-56\sqrt{3}}{97-56\sqrt{3}}=\frac{97-56\sqrt{3}}{{\left(97\right)}^2-{\left(56\sqrt{3}\right)}^2}=\frac{97-56\sqrt{3}}{9409-9408}=\frac{97-56\sqrt{3}}{1}=97-56\sqrt{3}$.

Therefore, $x^2+\frac{1}{x^2}$ can be computed as:

$x^2+\frac{1}{x^2}=97+56\sqrt{3}+97-56\sqrt{3}=97+97=194$.

Hence, $x^2+\frac{1}{x^2}=194$.

Therefore, option $B$ is correct.