Question

In each of the following determine rational numbers a and b :

(i) $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b\sqrt{3}$

(ii) $\frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b}$

(iii) $\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}$

(iv) $\frac{5+3\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$

(v) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b\sqrt{77}$

(vi) $\frac{4+3\sqrt{5}}{4-3\sqrt{5}}=a+b\sqrt{5}$

Answer 1

(i) $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\left(Rationalisingthedenominator\right)=\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}{)}^2-(1{)}^2}=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}Now,2-\sqrt{3}=a-b\sqrt{3}Comparingwegeta=2,b=1$

(ii) $\frac{4+\sqrt{2}}{2+\sqrt{2}}=\frac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}\left(Rationalisingthedenominator\right)=\frac{8-4\sqrt{2}+2\sqrt{2}-2}{(2{)}^2-(\sqrt{2}{)}^2}=\frac{6-2\sqrt{2}}{4-2}=\frac{6-2\sqrt{2}}{2}=3-\sqrt{2}Now,3-\sqrt{2}=a-\sqrt{b}Comparing,weget,a=3,b=2$

(iii) $\frac{3+\sqrt{2}}{3-\sqrt{2}}=\frac{(3+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}\left(Rationalisingthedenominator\right)=\frac{(3+\sqrt{2}{)}^2}{(3{)}^2-(\sqrt{2}{)}^2}=\frac{9+2+2\times 3\sqrt{2}}{9-2}=\frac{11+6\sqrt{2}}{7}=\frac{11}{7}+\frac{6}{7}\sqrt{2}Now,\frac{11}{7}+\frac{6}{7}\sqrt{2}=a+b\sqrt{2}Comparing,wegeta=\frac{11}{7},b=\frac{6}{7}$

(iv) $\frac{5+3\sqrt{3}}{7+4\sqrt{3}}=\frac{(5+3\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}\left(Rationalisingthedenominator\right)=\frac{35-20\sqrt{3}+21\sqrt{3}-12\times 3}{(7{)}^2-(4\sqrt{3}{)}^2}=\frac{35+\sqrt{3}-36}{49-48}=\frac{\sqrt{3}-1}{1}=-1+\sqrt{3}Now,-1+\sqrt{3}=a+b\sqrt{3}\therefore Comparing,wegeta=-1,b=1$

(v) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=\frac{(\sqrt{11}-\sqrt{7})(\sqrt{11}-\sqrt{7})}{(\sqrt{11}+\sqrt{7})(\sqrt{11}-\sqrt{7})}\left(Rationalisingthedenominator\right)=\frac{(\sqrt{11}-\sqrt{7}{)}^2}{(\sqrt{11}{)}^2-(\sqrt{7}{)}^2}=\frac{11+7-2\times \sqrt{11}\times \sqrt{7}}{11-7}=\frac{18-2\sqrt{77}}{4}=\frac{9-\sqrt{77}}{2}\therefore \frac{9}{2}-\frac{1}{2}\sqrt{77}=a-b\sqrt{77}\therefore Comparing,wegeta=\frac{9}{2},b=\frac{1}{2}$

(vi) $\frac{4+3\sqrt{5}}{4-3\sqrt{5}}=\frac{(4+3\sqrt{5})(4+3\sqrt{5})}{(4-3\sqrt{5})(4+3\sqrt{5})}\left(rationalisingthedenominator\right)=\frac{(4+3\sqrt{5}{)}^2}{(4{)}^2-(3\sqrt{5}{)}^2}=\frac{16+45+24\sqrt{5}}{16-45}=\frac{61+24\sqrt{5}}{-29}\therefore \frac{61+24\sqrt{5}}{-29}=a+b\sqrt{5}\Rightarrow \frac{-61}{29}-\frac{24}{29}\sqrt{5}=a+b\sqrt{5}Comparing,wegeta=\frac{-61}{29},b=\frac{-24}{29}$