document.write( "Question 292392: Rationalize the denominator. Assume that all variables represent positive real numbers and that the denominator is not zero.\r
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\n" ); document.write( "9- (3 under a radical)\r
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\n" ); document.write( "\n" ); document.write( "So it's 7 divided by 9 minus the square root of 3.
\n" ); document.write( "I think. I'm trying to write this so it makes sense.
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Algebra.Com's Answer #211212 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Here's how to write it so it makes sense:\r
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\n" ); document.write( "\n" ); document.write( "7/(9-sqrt(3))\r
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\n" ); document.write( "\n" ); document.write( "Anyone can read that to mean\r
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\n" ); document.write( "\n" ); document.write( "First review my lesson on Rationalizing Denominators: http://www.algebra.com/algebra/homework/Radicals/rationalizingdenominators1.lesson\r
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\n" ); document.write( "\n" ); document.write( "Then you need the conjugate of the denominator. If you have a binomial expression, such as , then its conjugate is \r
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\n" ); document.write( "\n" ); document.write( "So the conjugate of your denominator is \r
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\n" ); document.write( "\n" ); document.write( "Now multiply your expression by 1 in the form of the conjugate divided by itself.\r
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\n" ); document.write( "\n" ); document.write( "Remember that the product of a binomial and its conjugate is the difference of two squares:\r
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\n" ); document.write( "\n" ); document.write( "Yes, it's ugly, but at least there is no radical in the denominator.\r
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