document.write( "Question 149055: Hi. I'm an adult musician who sometimes uses mathy stuff to generate musical structures. I took algebra a long, long time ago and have forgotten most of it. A few years ago I stumbled upon this idea: for any odd number, the sum of the integers below that number is divisible by the odd number. For example, if I take 3, the sum of the integers below 3 -- 1 + 2 = 3, which is divisible by 3. Or 5: 1+2+3+4 = 10, which is divisible by 5. I got a friend to help me with an equation that would help me figure out the sum, so if I choose a number like 83, I don't have to spend 5 minutes punching 1+2+3 and so on to figure out the sum. That equation was x(x-1)/2 = the sum of the integers below x, when x is an odd number. \r
\n" ); document.write( "\n" ); document.write( "Now, for the first time I'm trying to run this backwards and I can't remember how to solve for it. In other words, let's call \"the sum of the integers below x,\" y. I know what y is, but I can't figure out x. For example, I've got 368 beats that I want to group in these different cycles. So I know that x(x-1)/2 = 368. I cross multiply the denominator and I get x(x-1) = 736. Then I multiply the x through and I get x^2 - x = 736. And then I'm stuck and don't know what to do to solve for x. I'm sure I'm doing something wrong or have forgotten some important step, because it's been forever since I've had to think about this stuff. Your help would be greatly appreciated.\r
\n" ); document.write( "\n" ); document.write( "cheers,
\n" ); document.write( "Andrea
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Algebra.Com's Answer #109348 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "\"x%5E2-x=736\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-x-736=0\" Get all terms to the left side.\r
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\n" ); document.write( "\n" ); document.write( "Notice we have a quadratic equation in the form of \"ax%5E2%2Bbx%2Bc\" where \"a=1\", \"b=-1\", and \"c=-736\"\r
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\n" ); document.write( "\n" ); document.write( "Let's use the quadratic formula to solve for x\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29\" Start with the quadratic formula\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%28-%28-1%29+%2B-+sqrt%28+%28-1%29%5E2-4%281%29%28-736%29+%29%29%2F%282%281%29%29\" Plug in \"a=1\", \"b=-1\", and \"c=-736\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+%28-1%29%5E2-4%281%29%28-736%29+%29%29%2F%282%281%29%29\" Negate \"-1\" to get \"1\". \r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+1-4%281%29%28-736%29+%29%29%2F%282%281%29%29\" Square \"-1\" to get \"1\". \r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+1--2944+%29%29%2F%282%281%29%29\" Multiply \"4%281%29%28-736%29\" to get \"-2944\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+1%2B2944+%29%29%2F%282%281%29%29\" Rewrite \"sqrt%281--2944%29\" as \"sqrt%281%2B2944%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+2945+%29%29%2F%282%281%29%29\" Add \"1\" to \"2944\" to get \"2945\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+2945+%29%29%2F%282%29\" Multiply \"2\" and \"1\" to get \"2\". \r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281%2Bsqrt%282945%29%29%2F%282%29\" or \"x+=+%281-sqrt%282945%29%29%2F%282%29\" Break up the expression. \r
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\n" ); document.write( "\n" ); document.write( "So our answers are \"x+=+%281%2Bsqrt%282945%29%29%2F%282%29\" or \"x+=+%281-sqrt%282945%29%29%2F%282%29\" \r
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\n" ); document.write( "\n" ); document.write( "which approximate to \"x=27.634\" or \"x=-26.634\" \r
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\n" ); document.write( "\n" ); document.write( "I'm not sure what you're looking for, but if you aren't expecting to get an irrational answer, then try \"y=378\"
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