document.write( "Question 45930: 
add or subtract as indicated \r\n" );
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document.write( "2Ö5 + 5Ö125
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Algebra.Com's Answer #30515 by AnlytcPhil(1806)\"\" \"About 
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add or subtract as indicated \r\n" );
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document.write( "2Ö5 + 5Ö125\r\n" );
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document.write( "You can't break 5 down because it's prime, but you can \r\n" );
document.write( "break down 125 into primes\r\n" );
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document.write( "125 = 5·25 = 5·5·5, so substitute 5·5·5 for 125\r\n" );
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document.write( "2Ö5 + 5Ö5·5·5\r\n" );
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document.write( "Since it's a square root we group like factors into \r\n" );
document.write( "twos. (If it were a cube root you would group into \r\n" );
document.write( "threes)\r\n" );
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document.write( "2Ö5 + 5Ö(5·5)·5\r\n" );
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document.write( "Since (5·5) is the same as 5², so we can substitute\r\n" );
document.write( "5² for (5·5) \r\n" );
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document.write( "2V5 + 5Ö5²·5\r\n" );
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document.write( "Now we can take individual square roots of the factors\r\n" );
document.write( "under the second radical\r\n" );
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document.write( "2Ö5 + 5Ö5²·Ö5\r\n" );
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document.write( "Taking the square root of '5 squared' just takes \r\n" );
document.write( "away the square and the radical, and you just have 5:\r\n" );
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document.write( "2Ö5 + 5·5·Ö5\r\n" );
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document.write( "2Ö5 + 25Ö5\r\n" );
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document.write( "Now you can factor out the Ö5\r\n" );
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document.write( "Ö5(2 + 25)\r\n" );
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document.write( "2 plus 25 is 27\r\n" );
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document.write( "Ö5(27)\r\n" );
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document.write( "Write the 27 first\r\n" );
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document.write( "27Ö5\r\n" );
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document.write( "Edwin
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