Question

Which option is closest to $\sqrt{65}-\sqrt{63}$?

• A. $0.125$
• B. $0.250$
• C. $0.146$
• D. $0.154$

Answer 1

The correct option is A $0.125$

$\sqrt{65}=(1+64{)}^{1/2}$ & $\sqrt{63}=(64-1{)}^{1/2}$

$\therefore \sqrt{65}-\sqrt{63}=(64+1{)}^{1/2}-(64-1{)}^{1/2}=8[{(1+\frac{1}{64})}^{1/2}-{(1-\frac{1}{64})}^{1/2}]$

Using Binomial Theorem $(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2-\mathrm{1.....};x<1$

${(1+\frac{1}{64})}^{1/2}=1+(\frac{1}{2}\times \frac{1}{64})-\frac{\frac{1}{4}\times {\left(\frac{1}{64}\right)}^2}{2!}+\frac{\frac{1}{2}\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)}{3!}{\left(\frac{1}{64}\right)}^3+...$

${(1-\frac{1}{64})}^{1/2}=1+\frac{1}{2}\left(\frac{1}{64}\right)+\frac{\frac{1}{2}-{\left(\frac{-1}{64}\right)}^2}{2!}+\frac{\frac{1}{2}\left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right){\left(\frac{-1}{64}\right)}^3}{3!}+....$

$\Rightarrow {(1+\frac{1}{64})}^{1/2}-{(1-\frac{1}{64})}^{1/2}=2\frac{1}{2}\left(\frac{1}{64}\right)+\frac{2{\left(\frac{1}{2}\right)}^2\left(\frac{3}{2}\right){\left(\frac{1}{64}\right)}^3}{3!}+...$

$\approx \frac{1}{64}$

$\Rightarrow \sqrt{65}-\sqrt{63}\approx \frac{8}{64}\approx 0.125$