Question

$\text{If}{x}^2=7-\sqrt{48}$, then $x$ = _______________

• A. $2+\sqrt{3}$
• B. $2-\sqrt{3}$
• C. $2\sqrt{3}$
• D. 4

Answer 1

The correct option is B

$2-\sqrt{3}$

$x^2=7-\sqrt{48}$

= $7-\sqrt{2^2\times 2^2\times 3}$

= $4+3-2\left(2\right)\left(\sqrt{3}\right)$

= $(2{)}^2+(\sqrt{3}{)}^2-2\left(2\right)\left(\sqrt{3}\right)$

=$(2{)}^2-2(2\left)\right(\sqrt{3})+(\sqrt{3}{)}^2$

=$(2-\sqrt{3}{)}^2$

$\therefore x=2-\sqrt{3}$