Question

If $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236$and $\sqrt{6}=2.449$, find the value of $\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}$

• A. $14.268$
• B. $18.428$
• C. $14.629$
• D. $14.662$

Answer 1

The correct option is A

$14.268$

Simplification

$\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}$ [Rationalization]

$$\begin{gathered} =\frac{{\left(2+\sqrt{3}\right)}^2}{2^2-{\sqrt{3}}^2}+\frac{{\left(2-\sqrt{3}\right)}^2}{2^2-{\sqrt{3}}^2}+\frac{{\left(\sqrt{3}-1\right)}^2}{{\left(\sqrt{3}\right)}^2-1^2} \\ =\frac{4+3+4\sqrt{3}}{4-3}+\frac{4+3-4\sqrt{3}}{4-3}+\frac{3+1-2\sqrt{3}}{3-1} \end{gathered}$$ $\left[{\left(a\pm b\right)}^2=a^2+b^2\pm 2ab,\left(a-b\right)\left(a+b\right)=a^2-b^2\right]$

$$\begin{gathered} =7+4\sqrt{3}+7-4\sqrt{3}+\frac{4-2\sqrt{3}}{2} \\ =14+2-\sqrt{3}=16-1.732=14.268 \end{gathered}$$

Therefore, $\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}$ is equal to $14.268$

Hence, option (A) is correct.