Answer:
Step-by-step explanation:
Given expressions are,
[tex]p=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}}[/tex] and [tex]q=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]
Remove the radicals from the denominator from both the expressions.
[tex]p=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}} \times \frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]
[tex]=\frac{(\sqrt{10}-\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}[/tex]
[tex]=\frac{(\sqrt{10}-\sqrt{5})^2}{5}[/tex]
[tex]\sqrt{p}=\sqrt{\frac{(\sqrt{10}-\sqrt{5})^2}{5}}[/tex]
[tex]=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{5}}[/tex]
[tex]q=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]
[tex]=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}\times \frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}[/tex]
[tex]=\frac{(\sqrt{10}+\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}[/tex]
[tex]=\frac{(\sqrt{10}+\sqrt{5})^2}{5}[/tex]
[tex]\sqrt{q}=\sqrt{\frac{(\sqrt{10}+\sqrt{5})^2}{5}}[/tex]
[tex]=\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}}[/tex]
[tex]\sqrt{q}-\sqrt{p}-2\sqrt{pq}=\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}}-\frac{(\sqrt{10}-\sqrt{5})}{\sqrt{5}}-2(\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}})(\frac{(\sqrt{10}-\sqrt{5})}{\sqrt{5}})[/tex]
[tex]=\frac{1}{\sqrt{5}}(\sqrt{10}+\sqrt{5}-\sqrt{10}+\sqrt{5})-\frac{2}{5}[(\sqrt{10})^2-(\sqrt{5})^2)][/tex]
[tex]=\frac{1}{\sqrt{5}}(2\sqrt{5})-\frac{2}{5}(10-5)[/tex]
[tex]=2-2[/tex]
[tex]=0[/tex]