Question

If 3+root2/ 3-root2 = a+b root2 , find the values a and b

Answer 1

Solving LHS:

$\small{=>\frac{3+\sqrt2}{3-\sqrt2}}$

Divide and multiply by $\small{\frac{3+\sqrt2}{3+\sqrt2}}$

$\small{=>\frac{3+\sqrt2}{3-\sqrt2}\times \frac{3+\sqrt2}{3+\sqrt2} }\\\\\small{=> \frac{(3+\sqrt2)^2}{(3+\sqrt2)(3-\sqrt2)} } \\\\\small{=>\frac{3^2+(\sqrt2)^2+2(3)(\sqrt2)}{3^2 - (\sqrt2)^2} }\\\\\small{=>\frac{9+2+6\sqrt2}{9-2}}\\\\\small{=>\frac{11+6\sqrt2}{7}}$

$=>\frac{11}{7}+\small{\frac{6\sqrt2}{7}}$

Matching the values

Then $\frac{6\sqrt2}{7}$ =b√2 → (6/7) = b

→ 11/7 =a

Answer 2

Answer:

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

Step-by-step explanation:

Given,

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

To find,

The value of 'a' and 'b'

Recall the identities

1. (a+b)² = a² +b² +2ab
2. (a+b)(a-b) =a² - b²

Solution:

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

First, we need to rationalize the denominator. The rationalizing factor is 3+√2

Multiply the numerator and denominator with the rationalizing factor, and we get

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } \ X \frac{3 + \sqrt{2} }{3 + \sqrt{2} }$ = $\frac{(3+ \sqrt{2})^2 }{(3-\sqrt{2})(3+\sqrt{2}) }$

Applying the identities (1) and (2)

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

Hence we have,

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} }$ =  a+b√2 = $\frac{11}{7} + \frac{6\sqrt{2} }{7}$

Comparing the terms we get,

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

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