Question

Find the square root of

(i) $\sqrt{\frac{2+\sqrt{3}}{2}}$

(ii) $\sqrt{4+\sqrt{7}}$

(iii) $\sqrt{7+3\sqrt{5}}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\sqrt{\frac{2+\sqrt{3}}{2}} \\ =\sqrt{1+\frac{\sqrt{3}}{2}} \\ =\sqrt{\frac{1}{4}+\frac{3}{4}+\frac{\sqrt{3}}{2}} \\ =\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2+2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}} \\ =\sqrt{{\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)}^2} \\ =\frac{1}{2}+\frac{\sqrt{3}}{2} \\ =\frac{1+\sqrt{3}}{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\sqrt{4+\sqrt{7}} \\ =\sqrt{\frac{1}{2}+\frac{7}{2}+\sqrt{7}} \\ =\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{\sqrt{7}}{\sqrt{2}}\right)}^2+2\times \frac{1}{\sqrt{2}}\times \frac{\sqrt{7}}{\sqrt{2}}} \\ =\sqrt{{\left(\frac{1}{\sqrt{2}}+\frac{\sqrt{7}}{\sqrt{2}}\right)}^2} \\ =\frac{1}{\sqrt{2}}+\frac{\sqrt{7}}{\sqrt{2}} \\ =\frac{1+\sqrt{7}}{\sqrt{2}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\sqrt{7+3\sqrt{5}} \\ =\sqrt{\frac{9}{2}+\frac{5}{2}+\sqrt{7}} \\ =\sqrt{{\left(\frac{3}{\sqrt{2}}\right)}^2+{\left(\frac{\sqrt{5}}{\sqrt{2}}\right)}^2+2\times \frac{3}{\sqrt{2}}\times \frac{\sqrt{5}}{\sqrt{2}}} \\ =\sqrt{{\left(\frac{3}{\sqrt{2}}+\frac{\sqrt{5}}{\sqrt{2}}\right)}^2} \\ =\frac{3}{\sqrt{2}}+\frac{\sqrt{5}}{\sqrt{2}} \\ =\frac{3+\sqrt{5}}{\sqrt{2}} \end{gathered}$$