document.write( "Question 276999: Put the equation 5x^2-10x-2 in turning point form a(x-h)^2+k, by completing the square. \n" ); document.write( "
Algebra.Com's Answer #201749 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "\"5x%5E2-10x-2\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "\"5%28x%5E2-2x-2%2F5%29\" Factor out the \"x%5E2\" coefficient \"5\". 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( "\"5%28x%5E2-2x%2Bhighlight%281-1%29-2%2F5%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( "\"5%28%28x%5E2-2x%2B1%29-1-2%2F5%29\" Group the first three terms.\r
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\n" ); document.write( "\n" ); document.write( "\"5%28%28x-1%29%5E2-1-2%2F5%29\" Factor \"x%5E2-2x%2B1\" to get \"%28x-1%29%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "\"5%28%28x-1%29%5E2-7%2F5%29\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"5%28x-1%29%5E2%2B5%28-7%2F5%29\" Distribute.\r
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\n" ); document.write( "\n" ); document.write( "\"5%28x-1%29%5E2-7\" Multiply.\r
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\n" ); document.write( "\n" ); document.write( "So after completing the square, \"5x%5E2-10x-2\" transforms to \"5%28x-1%29%5E2-7\". So \"5x%5E2-10x-2=5%28x-1%29%5E2-7\".\r
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