Question

If $\sqrt{2}$=1.414, then the square root of $\frac{\sqrt{2}-1}{\sqrt{2}+1}$ nearest to

• A. 0.172
• B. 0.414
• C. 0.586
• D. 1.414

Answer 1

The correct option is B 0.414

$\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{(\sqrt{2}-1{)}^2}{(\sqrt{2}+1)(\sqrt{2}-1)}$

$=\frac{2+1-2\sqrt{2}}{(\sqrt{2}{)}^2-1^2}$

$=\frac{3-2\sqrt{2}}{2-1}$

$=3-2\sqrt{2}$

$\therefore \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\sqrt{3-2\sqrt{2}}$

$=\sqrt{2+1-2\sqrt{2}}$

$=\sqrt{(\sqrt{2}{)}^2+1^2-2\sqrt{2}}$

$=\sqrt{(\sqrt{2}-1{)}^2}$

$=\sqrt{2}-1$

$=1.414-1$

$=0.414$