Question

3+√2\3-√2 =a+b√3. say me plzzzzz​

Answer 1

Answer:

$\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}\\\\\frac{(3+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}=a+b\sqrt{2}\\\\\frac{(3)^{2}+(\sqrt{2})^{2}+2(3)(\sqrt{2})}{(3)^{2}-(\sqrt{2}^{2})}=a+b\sqrt{2}\\\\\frac{9+2+6\sqrt{2}}{9-2}=a+b\sqrt{2}\\\\\frac{11+6\sqrt{2}}{7}=a+b\sqrt{2}\\\\\frac{11}{7}+\frac{6\sqrt{2}}{7}=a+b\sqrt{2}$

therefore,

$a=\frac{11}{7}$

$b\sqrt{2}=\frac{6\sqrt{2}}{7}\\\\b=\frac{6}{7}$

Answer 2

Answer:

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } = \frac{(3 + \sqrt{2}) {}^{2} }{9 - 2} = \frac{9 + 2 + 6 \sqrt{2} }{7} = \frac{11}{7} + \frac{6 \sqrt{2} }{7}$

So,

$a = \frac{11}{7} \: \: and \: \: b = \frac{6}{7}$