document.write( "Question 112305: y^2-12y+36=49\r
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Algebra.Com's Answer #81928 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"y%5E2-12y%2B36=49\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "\"y%5E2-12y-13=0\" Subtract 49 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"%28y-13%29%28y%2B1%29=0\" Factor the left side (note: if you need help with factoring, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "Now set each factor equal to zero:\r
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\n" ); document.write( "\n" ); document.write( "\"y-13=0\" or \"y%2B1=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"y=13\" or \"y=-1\" Now solve for y in each case\r
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\n" ); document.write( "\n" ); document.write( "So our solutions are \"y=13\" or \"y=-1\"\r
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\n" ); document.write( "\n" ); document.write( "Notice if we graph \"y=x%5E2-12x-13\" (just replace y with x) we get\r
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\n" ); document.write( "\n" ); document.write( "\"+graph%28500%2C500%2C-10%2C15%2C-10%2C10%2C+x%5E2-12x-13%29+\"\r
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\n" ); document.write( "\n" ); document.write( "and we can see that the graph has roots at \"y=13\" and \"y=-1\", so this verifies our answer.\r
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\n" ); document.write( "\n" ); document.write( "\"c%5E2%2B16c%2B64=15\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "\"c%5E2%2B16c%2B49=0\" Subtract 15 from both sides\r
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\n" ); document.write( "\n" ); document.write( "Let's use the quadratic formula to solve for c:\r
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\n" ); document.write( "\n" ); document.write( "Starting with the general quadratic\r
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\n" ); document.write( "\n" ); document.write( "\"ac%5E2%2Bbc%2Bc=0\"\r
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\n" ); document.write( "\n" ); document.write( "the general solution using the quadratic equation is:\r
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\n" ); document.write( "\n" ); document.write( "\"c+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29\"\r
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\n" ); document.write( "\n" ); document.write( "So lets solve \"c%5E2%2B16%2Ac%2B49=0\" ( notice \"a=1\", \"b=16\", and \"c=49\")\r
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\n" ); document.write( "\n" ); document.write( "\"c+=+%28-16+%2B-+sqrt%28+%2816%29%5E2-4%2A1%2A49+%29%29%2F%282%2A1%29\" Plug in a=1, b=16, and c=49\r
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\n" ); document.write( "\n" ); document.write( "\"c+=+%28-16+%2B-+sqrt%28+256-4%2A1%2A49+%29%29%2F%282%2A1%29\" Square 16 to get 256 \r
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\n" ); document.write( "\n" ); document.write( "\"c+=+%28-16+%2B-+sqrt%28+256%2B-196+%29%29%2F%282%2A1%29\" Multiply \"-4%2A49%2A1\" to get \"-196\"\r
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\n" ); document.write( "\n" ); document.write( "\"c+=+%28-16+%2B-+sqrt%28+60+%29%29%2F%282%2A1%29\" Combine like terms in the radicand (everything under the square root)\r
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\n" ); document.write( "\n" ); document.write( "\"c+=+%28-16+%2B-+2%2Asqrt%2815%29%29%2F%282%2A1%29\" Simplify the square root (note: If you need help with simplifying the square root, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "\"c+=+%28-16+%2B-+2%2Asqrt%2815%29%29%2F2\" Multiply 2 and 1 to get 2\r
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\n" ); document.write( "\n" ); document.write( "So now the expression breaks down into two parts\r
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\n" ); document.write( "\n" ); document.write( "\"c+=+%28-16+%2B+2%2Asqrt%2815%29%29%2F2\" or \"c+=+%28-16+-+2%2Asqrt%2815%29%29%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "Now break up the fraction\r
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\n" ); document.write( "\n" ); document.write( "\"c=-16%2F2%2B2%2Asqrt%2815%29%2F2\" or \"c=-16%2F2-2%2Asqrt%2815%29%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "\"c=-8%2Bsqrt%2815%29\" or \"c=-8-sqrt%2815%29\"\r
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\n" ); document.write( "\n" ); document.write( "So these expressions approximate to\r
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\n" ); document.write( "\n" ); document.write( "\"c=-4.12701665379258\" or \"c=-11.8729833462074\"\r
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\n" ); document.write( "\n" ); document.write( "So our solutions are:\r
\n" ); document.write( "\n" ); document.write( "\"c=-4.12701665379258\" or \"c=-11.8729833462074\"\r
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\n" ); document.write( "\n" ); document.write( "Notice when we graph \"x%5E2%2B16%2Ax%2B49\" (just replace c with x), we get:\r
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\n" ); document.write( "\n" ); document.write( "when we use the root finder feature on a calculator, we find that \"x=-4.12701665379258\" and \"x=-11.8729833462074\".So this verifies our answer\r
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