Question

find 3+root 2/3-root 2=a+b root2

Answer 1

Answer:

a = $\frac{11}{7}$ ; b = $\frac{6}{7}$

Step-by-step explanation:

$\frac{3 + \sqrt{2}}{3 - \sqrt{2} } = a + b\sqrt{2}$

Rationalising $\frac{3 + \sqrt{2}}{3 - \sqrt{2} }$ gives :-

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

Comparing $\frac{11 + 6\sqrt{2} }{7}$ with $a + b\sqrt{2}$ gives

a = $\frac{11}{7}$ & b = $\frac{6}{7}$

Answer 2

Answer:

$\frac{11}{7}$ + $\frac{6}{7}$ $\sqrt{2}$

Step-by-step explanation:

Given

$\frac{3+\sqrt{2} }{3-\sqrt{2} }$

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

The conjugate of 3 - $\sqrt{2}$ is 3 + $\sqrt{2}$ , thus

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

= $\frac{9+6\sqrt{2}+2 }{9-2}$

= $\frac{11+6\sqrt{2} }{7}$

= $\frac{11}{7}$ + $\frac{6}{7}$ $\sqrt{2}$ ← in the form a + b$\sqrt{2}$

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