Mathematics

Rationalise:
(1)              4/(2+root3+root7)
(2)              4/(2root3+root5)


Answers

lyssie11234

4 years ago Comment

Best answer

[latex]\frac{4}{2+\sqrt3+\sqrt7}\cdot\frac{2-(\sqrt3+\sqrt7)}{2-(\sqrt3+\sqrt7)}=\frac{8-4\sqrt3-4\sqrt7}{2^2-(\sqrt3+\sqrt7)^2}=\frac{8-4\sqrt3-4\sqrt7}{4-3-2\sqrt{3\cdot7}-7}\\\\=\frac{8-4\sqrt3-4\sqrt7}{-6-2\sqrt{21}}=\frac{-2(2\sqrt3+2\sqrt7-4)}{-2(3+\sqrt{21})}=\frac{2\sqrt3+2\sqrt7-4}{3+\sqrt{21}}\cdot\frac{3-\sqrt{21}}{3-\sqrt{21}}\\\\=\frac{6\sqrt3-2\sqrt{63}+6\sqrt7-2\sqrt{147}-12+4\sqrt{21}}{3^2-(\sqrt{21})^2}=\frac{6\sqrt3-2\sqrt{9\cdot7}+6\sqrt7-2\sqrt{49\cdot3}-12+4\sqrt{21}}{9-21}[/latex]

[latex]=\frac{6\sqrt3-6\sqrt7+6\sqrt7-14\sqrt3-12+4\sqrt{21}}{-12}=\frac{-8\sqrt3+4\sqrt{21}-12}{-12}=\frac{-4(2\sqrt3-\sqrt{21}+3)}{-12}\\\\=\frac{2\sqrt3-\sqrt{21}+3}{3}[/latex]

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[latex]\frac{4}{2\sqrt3+\sqrt5}\cdot\frac{2\sqrt3-\sqrt5}{2\sqrt3-\sqrt5}=\frac{8\sqrt3-4\sqrt5}{(2\sqrt3)^2-(\sqrt5)^2}=\frac{8\sqrt3-4\sqrt5}{4\cdot3-5}=\frac{8\sqrt3-4\sqrt5}{12-5}\\\\=\frac{8\sqrt3-4\sqrt5}{7}[/latex]

rishimisra15

4 years ago Comment

[latex](1) \frac{4}{2+\sqrt{3} +\sqrt{7}} \\ \\ or, \frac{4}{2+\sqrt{3} +\sqrt{7}} * \frac{2 - \sqrt{3} -\sqrt{7}}{2-\sqrt{3}-\sqrt{7}} \\ \\ => \frac{ \sqrt[2]{3} - \sqrt{21}+3}{3} \\ \\ \\ (2) \frac{4}{\sqrt[2]{3} + \sqrt{5}} \\ \\ or, \frac{4}{\sqrt[2]{3} + \sqrt{5}} * \frac{\sqrt[2]{3}-\sqrt{5}}{\sqrt[2]{3}-\sqrt{5}} \\ \\ => \frac{\sqrt[8]{3}-\sqrt[4]{5}}{7} [/latex]