Respuesta :

Answer:

a = [tex]\frac{11}{7}[/tex] ; b = [tex]\frac{6}{7}[/tex]

Step-by-step explanation:

[tex]\frac{3 + \sqrt{2}}{3 - \sqrt{2} } = a + b\sqrt{2} \\\\[/tex]

Rationalising [tex]\frac{3 + \sqrt{2}}{3 - \sqrt{2} }[/tex] gives :-

[tex]\frac{3 + \sqrt{2}}{3 - \sqrt{2} } = \frac{(3 + \sqrt{2})(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2}) } = \frac{(3 + \sqrt{2})^2}{3^2 - (\sqrt{2})^2 } = \frac{11 +6\sqrt{2} }{7}[/tex]

Comparing [tex]\frac{11 + 6\sqrt{2} }{7}[/tex] with [tex]a + b\sqrt{2}[/tex] gives

a = [tex]\frac{11}{7}[/tex] & b = [tex]\frac{6}{7}[/tex]

Answer:

[tex]\frac{11}{7}[/tex] + [tex]\frac{6}{7}[/tex] [tex]\sqrt{2}[/tex]

Step-by-step explanation:

Given

[tex]\frac{3+\sqrt{2} }{3-\sqrt{2} }[/tex]

Multiply the numerator/ denominator by the conjugate of the denominator.

The conjugate of 3 - [tex]\sqrt{2}[/tex] is 3 + [tex]\sqrt{2}[/tex] , thus

= [tex]\frac{(3+\sqrt{2})(3+\sqrt{2}) }{(3-\sqrt{2})(3+\sqrt{2}) }[/tex] ← expand numerator/ denominator using FOIL

= [tex]\frac{9+6\sqrt{2}+2 }{9-2}[/tex]

= [tex]\frac{11+6\sqrt{2} }{7}[/tex]

= [tex]\frac{11}{7}[/tex] + [tex]\frac{6}{7}[/tex] [tex]\sqrt{2}[/tex] ← in the form a + b[tex]\sqrt{2}[/tex]

with a = [tex]\frac{11}{7}[/tex] and b = [tex]\frac{6}{7}[/tex]

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