document.write( "Question 1045796: Kindly help me solving it
\n" ); document.write( "If \"a=3%2B2sqrt2\" then the value of \"a%5E%281%2F2%29%2Ba%5E%28-1%2F2%29\" is?
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Algebra.Com's Answer #661298 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
Let \"a=3%2B2sqrt%282%29\".\r
\n" ); document.write( "\n" ); document.write( "Now \"%28a%5E%281%2F2%29+%2B+a%5E%28-1%2F2%29%29%5E2+=+a+%2B+2+%2B+a%5E%28-1%29\"\r
\n" ); document.write( "\n" ); document.write( "Since \"%283%2B2sqrt%282%29%29%283-2sqrt%282%29%29+=+1\", then \"a%5E%28-1%29+=+1%2Fa+=+3+-2sqrt%282%29\".\r
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\n" ); document.write( "\n" ); document.write( "===> .\r
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\n" ); document.write( "\n" ); document.write( "Because \"a%5E%281%2F2%29+%2B+a%5E%28-1%2F2%29+%3E+0\", we then get \r
\n" ); document.write( "\n" ); document.write( "\"a%5E%281%2F2%29+%2B+a%5E%28-1%2F2%29+=+2sqrt%282%29\",\r
\n" ); document.write( "\n" ); document.write( "after taking square roots of both sides.
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