Question

$\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$ is equal to

• A. $2$
• B. $1$
• C. $2\sqrt{2}$
• D. $\sqrt{6}$

Answer 1

Answer: (A)

To find, value of : $\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$
Let $x=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$
Squaring both sides:
$x^2$ = ${(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}})}^2$
$x^2=3+2\sqrt{2}+3-2\sqrt{2}-2(3+2\sqrt{2})(3-2\sqrt{2})$
$\Rightarrow x^2=4$
$\Rightarrow x=2$
Hence, option 'A' is correct.