document.write( "Question 249026: I am struggling with this one ....any help is much appreciated!\r
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\n" ); document.write( "\n" ); document.write( "Square root of 5 over the square root of 180\r
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Algebra.Com's Answer #181437 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Here's one way to do it\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%285%29%2Fsqrt%28180%29\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%285%2F180%29\" Combine the roots using the identity \"sqrt%28x%29%2Fsqrt%28y%29=sqrt%28x%2Fy%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%281%2F36%29\" Reduce the fraction.\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%281%29%2Fsqrt%2836%29\" Break up the fraction using the identity \"sqrt%28x%2Fy%29=sqrt%28x%29%2Fsqrt%28y%29\" (ie it's the reverse of the other identity)\r
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\n" ); document.write( "\n" ); document.write( "\"1%2Fsqrt%2836%29\" Take the square root of 1 to get 1.\r
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\n" ); document.write( "\n" ); document.write( "\"1%2F6\" Take the square root of 36 to get 6.\r
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\n" ); document.write( "\n" ); document.write( "So \"sqrt%285%29%2Fsqrt%28180%29=1%2F6\"\r
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\n" ); document.write( "\n" ); document.write( "Or...\r
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\n" ); document.write( "\n" ); document.write( "Here's another way to do it. This method is more common as it works better in general.\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%285%29%2Fsqrt%28180%29\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "The goal is to get the denominator equal to a rational number. To do this, we multiply the denominator by \"sqrt%28180%29\". We must do the same for the numerator as well.\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%285%29%2Asqrt%28180%29%29%2F%28sqrt%28180%29%2Asqrt%28180%29%29\" Multiply both the numerator and denominator by \"sqrt%28180%29\"\r
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\n" ); document.write( "\n" ); document.write( "Scratch Work
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\n" ); document.write( "\n" ); document.write( "Take note that \"sqrt%28180%29%2Asqrt%28180%29=sqrt%28180%2A180%29\" since \"sqrt%28x%2Ay%29=sqrt%28x%29%2Asqrt%28y%29\" (it's like those previous identities). Since \"sqrt%28180%29%2Asqrt%28180%29=sqrt%28180%2A180%29\", we can say that \"sqrt%28180%29%2Asqrt%28180%29=sqrt%28180%2A180%29=sqrt%28180%5E2%29=180\". \r
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\n" ); document.write( "\n" ); document.write( "For the last step of that equation, the square root 'undoes' the square. In other words, \"sqrt%28x%5E2%29=x\" where \"x%3E=0\"\r
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\n" ); document.write( "\n" ); document.write( "In the end, \"sqrt%28180%29%2Asqrt%28180%29=180\"\r
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\n" ); document.write( "\n" ); document.write( "Now back to the problem at hand....\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%285%29%2Asqrt%28180%29%29%2F180\" Multiply \"sqrt%28180%29\" with \"sqrt%28180%29\" to get 180 (see scratch work above).\r
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\n" ); document.write( "\n" ); document.write( "More Scratch Work
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\n" ); document.write( "\n" ); document.write( "Following the same idea, \"sqrt%285%29%2Asqrt%28180%29=sqrt%285%2A180%29\" due to the identity \"sqrt%28x%2Ay%29=sqrt%28x%29%2Asqrt%28y%29\". \r
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\n" ); document.write( "\n" ); document.write( "So \"sqrt%285%29%2Asqrt%28180%29=sqrt%285%2A180%29=sqrt%28900%29=sqrt%2830%5E2%29=30\"\r
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\n" ); document.write( "\n" ); document.write( "In the end, \"sqrt%285%29%2Asqrt%28180%29=30\"\r
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\n" ); document.write( "\n" ); document.write( "If you're not sure about your work, a calculator will come in handy to check your work.\r
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\n" ); document.write( "\n" ); document.write( "Now back to the problem at hand....\r
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\n" ); document.write( "\n" ); document.write( "\"30%2F180\" Multiply \"sqrt%285%29\" with \"sqrt%28180%29\" to get 30 (see scratch work above).\r
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\n" ); document.write( "\n" ); document.write( "\"1%2F6\" Reduce the fraction.\r
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\n" ); document.write( "\n" ); document.write( "So \"sqrt%285%29%2Fsqrt%28180%29=1%2F6\" regardless of the method used to get the answer.
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