document.write( "Question 1192822: Rationalize the denominator\r
\n" ); document.write( "\n" ); document.write( "sqrt(2x)/sqrt(x)-sqrt(10)
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Algebra.Com's Answer #824736 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "\"%28sqrt%282x%29%29%2F%28sqrt%28x%29-sqrt%2810%29%29\" is in the form \"A%2F%28B-C%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"A+=+sqrt%282x%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"B+=+sqrt%28x%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"C+=+sqrt%2810%29\"\r
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\n" ); document.write( "\n" ); document.write( "Multiply top and bottom by (B+C)
\n" ); document.write( "This way we have in the denominator, thereby rationalizing it (i.e. removing the square root terms from the denominator). Refer to the difference of squares rule.\r
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\n" ); document.write( "\n" ); document.write( "\"A%2F%28B-C%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28A%28B%2BC%29%29%2F%28%28B-C%29%28B%2BC%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28A%28B%2BC%29%29%2F%28B%5E2+-+C%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%282x%29%2A%28sqrt%28x%29%2Bsqrt%2810%29%29%29%2F%28x-10%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%282x%29%2Asqrt%28x%29%2Bsqrt%282x%29%2Asqrt%2810%29%29%2F%28x-10%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%282x%2Ax%29%2Bsqrt%282x%2A10%29%29%2F%28x-10%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%282x%5E2%29%2Bsqrt%2820x%29%29%2F%28x-10%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%28x%5E2%2A2%29%2Bsqrt%284%2A5x%29%29%2F%28x-10%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%28x%5E2%29%2Asqrt%282%29%2Bsqrt%284%29%2Asqrt%285x%29%29%2F%28x-10%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28x%2Asqrt%282%29%2B2%2Asqrt%285x%29%29%2F%28x-10%29\"\r
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\n" ); document.write( "\n" ); document.write( "Therefore,
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\n" ); document.write( "such that \"x+%3E=+0\" and \"x+%3C%3E+10\"
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