Question 16
If $\sqrt{2}=1.4142$... then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to

A) 2.4142
B) 5.8282
C) 0.4142
D) 0.1718

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

[inside the root, mulitplying numerator and denominator by $(\sqrt{2}-1)$]

[Using identify $(a-b)(a+b)=a^2-b^2$]

$=\sqrt{\frac{(\sqrt{2}-1{)}^2}{(\sqrt{2}{)}^2-(1{)}^2}}=\sqrt{\frac{(\sqrt{2}-1{)}^2}{2-1}}=\sqrt{2}-1=\left(\mathrm{1.4142...}\right)-1=0.4142$