document.write( "Question 19274: \"X%5E2-6x=16\" I'm supposed to solve this by completing the square. \n" ); document.write( "
Algebra.Com's Answer #9302 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
right, fairly easy concept here. As follows:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-6x=16\" Just look at the left hand side: \"x%5E2-6x\". We need to figure out which number to add/subtract such that we can then factorise it as \"x%2Ba%29%5E2\" or \"%28x-a%29%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "so, \"x%2Ba%29%5E2\" expanded would give \"x%5E2+%2B+2ax+%2B+a%5E2\" and the negative version would have -2ax instead.\r
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\n" ); document.write( "\n" ); document.write( "So, we have -6x which is the (-2ax) which means that a needs to be 3. \r
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\n" ); document.write( "\n" ); document.write( "So, the \"a%5E2\" term is \"3%5E2\" ---> 9\r
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\n" ); document.write( "\n" ); document.write( "So, we have to add 9...to both sides remember, to keep the equation unaltered... \"x%5E2-6x%2B9+=+25\"\r
\n" ); document.write( "\n" ); document.write( "So, \"x%5E2+-+6x+%2B+9\" can be written instantly as \"%28x-3%29%5E2\" --> the negative version, since the x-term was negative.\r
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\n" ); document.write( "\n" ); document.write( "and we have \"%28x-3%29%5E2+=+25\"
\n" ); document.write( "\"%28x-3%29+=+%2Bsqrt%2825%29\" OR \"%28x-3%29+=+-sqrt%2825%29\"\r
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\n" ); document.write( "\n" ); document.write( "--> x-3 = +5 OR x-3 = -5
\n" ); document.write( "--> x = 8 OR x = -2\r
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\n" ); document.write( "\n" ); document.write( "jon.
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