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Question
If m = `1/[ 3 - 2sqrt2 ] and n = 1/[ 3 + 2sqrt2 ],` find m2
Sum
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Solution
m = `1/[ 3 - 2sqrt2 ]`
m = `1/[ 3 - 2sqrt2 ] xx [ 3 + 2sqrt2 ]/[ 3 + 2sqrt2 ]`
m = `[ 3 + 2sqrt2 ]/[ (3)^2 - (2sqrt2)^2 ]`
m = `[ 3 + 2sqrt2 ]/[ 9 - 8 ]`
m = 3 +2√2
⇒ m2 = ( 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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Rationalisation of Surds
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