Question

If $x=7-4\sqrt{3}$, the value of $x^2+\frac{1}{x^2}$ will be

• A. $146$
• B. $148$
• C. $194$
• D. $196$

Answer 1

The correct option is C $194$

Given,
$x=7-4\sqrt{3}$
$\therefore \frac{1}{x}=\frac{1}{(7-4\sqrt{3})}\times \frac{7+4\sqrt{3}}{7+4\sqrt{3}}$
$=\frac{7+4\sqrt{3}}{(49-48)}$
$=7+4\sqrt{3}$
Now,
$x+\frac{1}{x}=(7-4\sqrt{3})+(7+4\sqrt{3})$
$=14$
$=>{(x+\frac{1}{x})}^2=(14{)}^2$
Using $(a+b{)}^2=a^2+b^2+2ab$ we get,
$=>x^2+\frac{1}{x^2}+2\left(x\right)\left(\frac{1}{x}\right)=196$
$=>x^2+\frac{1}{x^2}=196-2$
$\therefore x^2+\frac{1}{x^2}=194$