Question

If $a+b\sqrt{2} =\frac{3+\sqrt{2} }{3-\sqrt{2} }$, find the value of a and b

Answer 1

Answer:

Given

\dfrac{3}{2-\sqrt{2} } = a+b\sqrt{2}

LHS

= \dfrac{3}{2-\sqrt{2} }

= \dfrac{3}{2-\sqrt{2} } \times \dfrac{2+\sqrt{2} }{2+\sqrt{2} }

= \dfrac{6+3\sqrt{2} }{(2)^{2} -(\sqrt{2})^{2} }

= \dfrac{6+3\sqrt{2} }{4-2 }

= \dfrac{6+3\sqrt{2} }{2 }

= 2 + \dfrac{3\sqrt{2} }{2}

Therefore,

\boxed{\boxed{\bold{a=2}}}}}}}}}

\boxed{\boxed{\bold{b=\frac{3}{2} }}}}}