rationalize the denominator under root 3 minus 1 by under root 3 + 1 + 2root 2
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Hi ,
( √3 - 1 )/ ( √3 + 1 + 2√2 )
= [(√3-1)( 1+2√2-√3)]/[(1+2√2+√3)(1+2√2-√3)]
= [(√3+2√6-3-1-2√2+√3 )]/[(1+2√2)²-(√3)² ]
= [ -4 -2√2 +2√3 + 2√6 ]/[ 1 +8 +4√2 - 3 ]
= [ 2( -2 -√2 + √3 + √6 ]/[ 6 + 4√2)
= ( -2 - √2 + √3 + √6 ) / ( 3 + 2√2 )
= [(-2-√2+√3+√6)(3-2√2)]/[(3+2√2)(3-2√2)]
= [-6+4√2-3√2+4+3√3-2√6+3√6-2√3]/(9-8)
= -2 + √2 + √3 + √6
I hope this helps you.
: )
( √3 - 1 )/ ( √3 + 1 + 2√2 )
= [(√3-1)( 1+2√2-√3)]/[(1+2√2+√3)(1+2√2-√3)]
= [(√3+2√6-3-1-2√2+√3 )]/[(1+2√2)²-(√3)² ]
= [ -4 -2√2 +2√3 + 2√6 ]/[ 1 +8 +4√2 - 3 ]
= [ 2( -2 -√2 + √3 + √6 ]/[ 6 + 4√2)
= ( -2 - √2 + √3 + √6 ) / ( 3 + 2√2 )
= [(-2-√2+√3+√6)(3-2√2)]/[(3+2√2)(3-2√2)]
= [-6+4√2-3√2+4+3√3-2√6+3√6-2√3]/(9-8)
= -2 + √2 + √3 + √6
I hope this helps you.
: )
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