Question

3-2√2/3+2√2=a+b√2 find the value of a and b​

Answer 1

Hey there! Here is your answer.

$\frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } \\ = \frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ = \frac{ {(3 - 2 \sqrt{2)} }^{2} }{ {3}^{2} - ( {2 \sqrt{2)} }^{2} } \\ = \frac{ {3}^{2} - 2 \times 3 \times 2 \sqrt{2} {(2 \sqrt{2})}^{2} }{9 - 8} \\ = \frac{9 - 12 \sqrt{2} + 8}{ 1} \\ = 17 - 12 \sqrt{2} \\ a \: = 17 \\ b = 12$

Hope this answer will help you!

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Answer 2

solution = 3-

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

$2 \sqrt{2} - 2 \sqrt{2}$

cut

$3 \div 3 = a + b \sqrt{2}$

$b \sqrt{2} = a - 1$

$b = a - 1 \div \sqrt{2}$