Question

find the value of "a" and "b"​

Answer 1

Solve the given equation:

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

Rationalize it:

=> $\tt{\frac{3 + \sqrt{2}}{3 - 2\sqrt{2}} \times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} }$

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

=> $\tt{\frac{9 + 6\sqrt{2} + 3\sqrt{2} + 4}{9 - 8}}$

=> $\tt{13 + 9\sqrt{2}}$

Equate this to the RHS:

$\tt{13 + 9\sqrt{2} = a + b\sqrt{2}}$

Thus, we get:

a = 13

b = 9

Answer 2

Solutions:

(3 + √2) / (3 - 2√2) = a + b√2

(3 + √2) / (3 - 2√2) × (3 + 2√2) / (3 + 2√2)

(3 + 2√2) / (3 + 2√2) / (3)² - (2√2)²

(9 + 6√2 + 3√2 + 4) / (9 - 8) = (13 + 9√2)

a = 13

b = 9

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