Question

If $x=\sqrt{\frac{2+\sqrt{3}}{2-\surd 3}}$, then $x^2{(x-4)}^2=$

• A. $7$
• B. $4$
• C. $2$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for the correct option.

Step 1: Simplify $x$

$\begin{array}{rcl}x& =& \sqrt{\frac{2+\sqrt{3}}{2-\surd 3}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}}\\ & =& \sqrt{\frac{{\left(2+\sqrt{3}\right)}^2}{4-3}}\\ & =& \sqrt{{\left(2+\sqrt{3}\right)}^2}\\ & =& \left(2+\sqrt{3}\right)\end{array}$

Step 2: Find the value of $x^2{(x-4)}^2$

$\begin{array}{rcl}{x}^2{(x-4)}^2& =& {\left[x\left(x-4\right)\right]}^2\\ & =& {\left[\left(2+\sqrt{3}\right)\left(2+\sqrt{3}-4\right)\right]}^2\\ & =& {\left[\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)\right]}^2\\ & =& {\left[2\sqrt{3}-4+3-2\sqrt{3}\right]}^2\\ & =& {\left(-1\right)}^2\\ & =& 1\end{array}$

Hence, option D is correct.