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Question
If m = `1/[ 3 - 2sqrt2 ] and n = 1/[ 3 + 2sqrt2 ],` find n2
Solution
n = `1/[ 3 + 2sqrt2 ]`
n = `1/[ 3 + 2sqrt2 ] xx [ 3 - 2sqrt2 ]/[ 3 - 2sqrt2 ]`
n = `[ 3 - 2sqrt2 ]/[ (3)^2 - (2sqrt2)^2 ]`
n = `[ 3 - 2sqrt2 ]/[ 9 - 8 ]`
n = 3 - 2√2
⇒ n2 = ( 3 - 2√2)2
= (3)2 - 2 x 3 x 2√2 + (2√2)2
= 9 - 12√2 + 8
= 17 - 12√2
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