Question

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

Answer 1

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,

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

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

                                                       

Answer 2

Step-by-step explanation:

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