document.write( "Question 198574: Can someone explain how this problem works (if possible step by step so that i can understand it)
\n" ); document.write( "solve using completing the square, 3x^2-6x=27 the answer is x=1+sqrt(10), 1-sqrt(10), why?
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Algebra.Com's Answer #149094 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"3x%5E2-6x=27\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"3x%5E2-6x-27=0\" Subtract 27 from both sides.\r
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\n" ); document.write( "\n" ); document.write( "Now let's complete the square for the left side expression \"3x%5E2-6x-27\"\r
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\n" ); document.write( "\n" ); document.write( "\"3x%5E2-6x-27\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "\"3%28x%5E2-2x-9%29\" Factor out the \"x%5E2\" coefficient \"3\". This step is very important: the \"x%5E2\" coefficient must be equal to 1.\r
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\n" ); document.write( "\n" ); document.write( "Take half of the \"x\" coefficient \"-2\" to get \"-1\". In other words, \"%281%2F2%29%28-2%29=-1\".\r
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\n" ); document.write( "\n" ); document.write( "Now square \"-1\" to get \"1\". In other words, \"%28-1%29%5E2=%28-1%29%28-1%29=1\"\r
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\n" ); document.write( "\n" ); document.write( "\"3%28x%5E2-2x%2Bhighlight%281-1%29-9%29\" Now add and subtract \"1\" inside the parenthesis. Make sure to place this after the \"x\" term. Notice how \"1-1=0\". So the expression is not changed.\r
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\n" ); document.write( "\n" ); document.write( "\"3%28%28x%5E2-2x%2B1%29-1-9%29\" Group the first three terms.\r
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\n" ); document.write( "\n" ); document.write( "\"3%28%28x-1%29%5E2-1-9%29\" Factor \"x%5E2-2x%2B1\" to get \"%28x-1%29%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "\"3%28%28x-1%29%5E2-10%29\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"3%28x-1%29%5E2%2B3%28-10%29\" Distribute.\r
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\n" ); document.write( "\n" ); document.write( "\"3%28x-1%29%5E2-30\" Multiply.\r
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\n" ); document.write( "\n" ); document.write( "So after completing the square, \"3x%5E2-6x-27\" transforms to \"3%28x-1%29%5E2-30\". So \"3x%5E2-6x-27=3%28x-1%29%5E2-30\".\r
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\n" ); document.write( "\n" ); document.write( "So \"3x%5E2-6x-27=0\" is equivalent to \"3%28x-1%29%5E2-30=0\".\r
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\n" ); document.write( "\n" ); document.write( "Now let's solve \"3%28x-1%29%5E2-30=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"3%28x-1%29%5E2-30=0\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"3%28x-1%29%5E2=0%2B30\"Add \"30\" to both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"3%28x-1%29%5E2=30\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-1%29%5E2=%2830%29%2F%283%29\" Divide both sides by \"3\".\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-1%29%5E2=10\" Reduce.\r
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\n" ); document.write( "\n" ); document.write( "\"x-1=%22%22%2B-sqrt%2810%29\" Take the square root of both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"x-1=sqrt%2810%29\" or \"x-1=-sqrt%2810%29\" Break up the \"plus/minus\" to form two equations.\r
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\n" ); document.write( "\n" ); document.write( "\"x-1=sqrt%2810%29\" or \"x-1=-sqrt%2810%29\" Simplify the square root.\r
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\n" ); document.write( "\n" ); document.write( "\"x=1%2Bsqrt%2810%29\" or \"x=1-sqrt%2810%29\" Add \"1\" to both sides.\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the solutions are \"x=1%2Bsqrt%2810%29\" or \"x=1-sqrt%2810%29\".\r
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\n" ); document.write( "\n" ); document.write( "Note: you can use the quadratic formula to verify your answer.
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