Question

The value of $\frac{1}{\sqrt{3}+\sqrt{2}-1}$ on simplifying upto 3 decimal places, given that $\sqrt{2}=1.4142$ and $\sqrt{6}=2.4495$, is:

• A. $0.166$
• B. $0.366$
• C. $0.466$
• D. $0.566$

Answer 1

The correct option is C $0.466$

We will first rationalize the denominators of the given expression and then solve it as follows:

$\frac{1}{(\sqrt{3}+\sqrt{2})-1}\times \frac{(\sqrt{3}+\sqrt{2})+1}{(\sqrt{3}+\sqrt{2})+1}=\frac{(\sqrt{3}+\sqrt{2})+1}{{(\sqrt{3}+\sqrt{2})}^2-1^2}\{\because (a-b)(a+b)=a^2-b^2\}=\frac{(\sqrt{3}+\sqrt{2})+1}{[{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2+(2\times \sqrt{3}\times \sqrt{2})]-1}\{\because {(a+b)}^2=({a^2+b}^2+2ab)\}=\frac{(\sqrt{3}+\sqrt{2})+1}{[3+2+2\sqrt{6}]-1}$

$=\frac{(\sqrt{3}+\sqrt{2})+1}{5+2\sqrt{6}-1}=\frac{(\sqrt{3}+\sqrt{2})+1}{4+2\sqrt{6}}.......\left(1\right)$ .

It is given that the value of $\sqrt{2}=1.4142,$ $\sqrt{3}=1.73205$ and $\sqrt{6}=2.4495$, therefore we substitute these values in equation $\left(1\right)$ as shown below:

$\frac{(1.73205+1.4142)+1}{4+(2\times 2.4495)}=\frac{3.14625+1}{4+4.899}=\frac{4.14625}{8.899}=\mathrm{0.46592313....}$, when corrected to three decimal points.

Hence, $\frac{1}{(\sqrt{3}+\sqrt{2})-1}=0.466$.

Therefore, option $C$ is correct.