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Question Number 173869 by mnjuly1970 last updated on 20/Jul/22

$$$$$$If,{\left(\frac{\sqrt{10}+\sqrt{6}+2}{\sqrt{5}-\sqrt{3}+\sqrt{2}}\right)}^2=a+\sqrt{b}$$$$then.a,b=?$$$$$$

Commented by infinityaction last updated on 20/Jul/22

$$\frac{\sqrt{2}(\sqrt{5}+\sqrt{3}+\sqrt{2})}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=A$$$$\frac{\sqrt{2}(\sqrt{5}+\sqrt{3})(1+\frac{\sqrt{2}}{\sqrt{5}+\sqrt{3}})}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=A$$$$\frac{\sqrt{5}+\sqrt{3}(\sqrt{2}+\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})})}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=A$$$$\frac{\sqrt{5}+\sqrt{3}\cancel{(\sqrt{2}+\frac{2(\sqrt{5}-\sqrt{3})}{2})}}{\cancel{\sqrt{5}+\sqrt{2}-\sqrt{3}}}=A$$$$\sqrt{5}+\sqrt{3}=A$$$$A^2={(\sqrt{5}+\sqrt{3})}^2=a+\sqrt{b}$$$$a+\sqrt{b}=8+\sqrt{60}$$$$a=8,b=60$$

Commented by mnjuly1970 last updated on 20/Jul/22

$$thxalotsir$$

Answered by Rasheed.Sindhi last updated on 20/Jul/22

$$If{\left(\frac{\sqrt{10}+\sqrt{6}+2}{\sqrt{5}-\sqrt{3}+\sqrt{2}}\right)}^2=a+\sqrt{b}$$$${\left(\frac{\sqrt{10}+\sqrt{6}+2}{\sqrt{5}-\sqrt{3}+\sqrt{2}}\right)}^2$$$$=\frac{10+6+4+4\sqrt{15}+4\sqrt{6}+4\sqrt{10}}{5+3+2-2\sqrt{15}-2\sqrt{6}+2\sqrt{10}}$$$$=\frac{20+4\sqrt{15}+4\sqrt{6}+4\sqrt{10}}{10-2\sqrt{15}-2\sqrt{6}+2\sqrt{10}}$$$$=\frac{10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}}{5-\sqrt{15}-\sqrt{6}+\sqrt{10}}=\frac{p}{q}\left(say\right)$$$$\frac{10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}}{10-2\sqrt{15}-2\sqrt{6}+2\sqrt{10}}=\frac{p}{2q}$$$$\frac{(10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10})+(10-2\sqrt{15}-2\sqrt{6}+2\sqrt{10})}{(10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10})-(10-2\sqrt{15}-2\sqrt{6}+2\sqrt{10})}=\frac{p+2q}{p-2q}$$$$\frac{20+4\sqrt{10}}{4\sqrt{15}+4\sqrt{6}}=\frac{p+2q}{p-2q}$$$$\frac{5+\sqrt{10}}{\sqrt{15}+\sqrt{6}}\cdot \frac{\sqrt{15}-\sqrt{6}}{\sqrt{15}-\sqrt{6}}=\frac{p+2q}{p-2q}$$$$\frac{5\sqrt{15}-5\sqrt{6}+5\sqrt{6}-2\sqrt{15}}{15-6}=\frac{p+2q}{p-2q}$$$$\frac{3\sqrt{15}}{9}=\frac{p+2q}{p-2q}$$$$\frac{\sqrt{15}}{3}=\frac{p+2q}{p-2q}$$$$\frac{\sqrt{15}+3}{\sqrt{15}-3}=\frac{(p+2q)+(p-2q}{(p+2q)-(p-2q)}$$$$\frac{\sqrt{15}+3}{\sqrt{15}-3}=\frac{2p}{4q}=\frac{p}{2q}$$$$\frac{\sqrt{15}+3}{\sqrt{15}-3}\cdot \frac{\sqrt{15}+3}{\sqrt{15}+3}=\frac{p}{2q}$$$$\frac{{(\sqrt{15}+3)}^2}{15-9}=\frac{p}{2q}$$$$\frac{15+9+6\sqrt{15}}{6}=\frac{p}{2q}$$$$\frac{24+6\sqrt{15}}{3}=\frac{p}{q}$$$$=8+2\sqrt{15}=8+\sqrt{60}$$$$a=8,b=60$$

Answered by Rasheed.Sindhi last updated on 20/Jul/22

$$If{\left(\frac{\sqrt{10}+\sqrt{6}+2}{\sqrt{5}-\sqrt{3}+\sqrt{2}}\right)}^2=a+\sqrt{b}$$$$then.a,b=?$$$$\frac{\sqrt{10}+\sqrt{6}+2}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=\frac{\sqrt{2}(\sqrt{5}+\sqrt{3}+\sqrt{2})}{\sqrt{5}-\sqrt{3}+\sqrt{2}}$$$${=}^{}\frac{\sqrt{2}(\sqrt{5}+\sqrt{2}+\sqrt{3})}{\sqrt{5}+\sqrt{2}-\sqrt{3}}\cdot \frac{(\sqrt{5}+\sqrt{2}+\sqrt{3})}{(\sqrt{5}+\sqrt{2}+\sqrt{3})}$$$$=\frac{\sqrt{2}{(\sqrt{5}+\sqrt{2}+\sqrt{3})}^2}{{(\sqrt{5}+\sqrt{2})}^2-{\left(\sqrt{3}\right)}^2}$$$$=\frac{\sqrt{2}(5+2+3+2\sqrt{10}+2\sqrt{6}+2\sqrt{15})}{5+2+2\sqrt{10}-3}$$$$=\frac{\sqrt{2}(10+2\sqrt{10}+2\sqrt{6}+2\sqrt{15})}{4+2\sqrt{10}}$$$$=\frac{\sqrt{2}(5+\sqrt{10}+\sqrt{6}+\sqrt{15})}{2+\sqrt{10}}\cdot \frac{2-\sqrt{10}}{2-\sqrt{10}}$$$$=\frac{\sqrt{2}(10+2\sqrt{10}+2\sqrt{6}+2\sqrt{15}-5\sqrt{10}-10-\sqrt{60}-\sqrt{150})}{4-10}$$$$=\frac{\sqrt{2}(\cancel{10}-3\sqrt{10}-\cancel{10}-3\sqrt{6})}{-6}$$$$=\frac{\sqrt{2}(\sqrt{10}+\sqrt{6})}{2}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{2}}=\sqrt{5}+\sqrt{3}$$$$▸{\left(\frac{\sqrt{10}+\sqrt{6}+2}{\sqrt{5}-\sqrt{3}+\sqrt{2}}\right)}^2={(\sqrt{5}+\sqrt{3})}^2$$$$=5+3+2\sqrt{15}=8+2\sqrt{15}=8+\sqrt{60}$$$$a=8,b=60$$

Commented by mnjuly1970 last updated on 20/Jul/22

$$verynicesir$$$$thankyousomuch$$

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