document.write( "Question 1172153: Find the quadratic equation in x with the given roots 1±3i√5 / 2 <- (it is a fraction). Show clear solution thanksss! \n" ); document.write( "
Algebra.Com's Answer #797076 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Use the zero product property to help go from the given roots to a quadratic polynomial.\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281%2B3i%2Asqrt%285%29%29%2F2\" or \"x+=+%281-3i%2Asqrt%285%29%29%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "\"2x+=+1%2B3i%2Asqrt%285%29\" or \"2x+=+1-3i%2Asqrt%285%29\" Multiply both sides by 2\r
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\n" ); document.write( "\n" ); document.write( "\"2x-1+=+3i%2Asqrt%285%29\" or \"2x-1+=+-3i%2Asqrt%285%29\" Subtract 1 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"%282x-1%29%5E2+=+%283i%2Asqrt%285%29%29%5E2\" or \"%282x-1%29%5E2+=+%28-3i%2Asqrt%285%29%29%5E2\" Square both sides\r
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\n" ); document.write( "\n" ); document.write( "\"%282x-1%29%5E2+=+9i%5E2%2A5\"\r
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\n" ); document.write( "\n" ); document.write( "\"%282x-1%29%5E2+=+9%28-1%29%2A5\" Use i^2 = -1\r
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\n" ); document.write( "\n" ); document.write( "\"%282x-1%29%5E2+=+-45\"\r
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\n" ); document.write( "\n" ); document.write( "\"%282x-1%29%5E2%2B45+=+0\"\r
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\n" ); document.write( "\n" ); document.write( "\"4x%5E2-4x%2B1%2B45+=+0\" FOIL\r
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\n" ); document.write( "\n" ); document.write( "\"4x%5E2-4x%2B46+=+0\" \r
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\n" ); document.write( "\n" ); document.write( "\"2%282x%5E2-2x%2B23%29+=+0\" \r
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\n" ); document.write( "\n" ); document.write( "\"2x%5E2-2x%2B23+=+0%2F2\"\r
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\n" ); document.write( "\"2x%5E2-2x%2B23+=+0\" This is the final answer.\r
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\n" ); document.write( "\n" ); document.write( "We can use the quadratic formula to confirm this\r
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\n" ); document.write( "\n" ); document.write( "For \"2x%5E2-2x%2B23+=+0\", we have a = 2, b = -2, c = 23\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29\" or \"x+=+%28-b-sqrt%28b%5E2-4ac%29%29%2F%282a%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%28-%28-2%29%2Bsqrt%28%28-2%29%5E2-4%282%29%2823%29%29%29%2F%282%282%29%29\" or \"x+=+%28-%28-2%29-sqrt%28%28-2%29%5E2-4%282%29%2823%29%29%29%2F%282%282%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%282%2Bsqrt%28-180%29%29%2F%282%282%29%29\" or \"x+=+%282-sqrt%28-180%29%29%2F%282%282%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%282%2Bsqrt%2836%2A%28-1%29%2A5%29%29%2F%282%282%29%29\" or \"x+=+%282-sqrt%2836%2A%28-1%29%2A5%29%29%2F%282%282%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%282%2Bsqrt%2836%29%2Asqrt%28-1%29%2Asqrt%285%29%29%2F%282%282%29%29\" or \"x+=+%282-sqrt%2836%29%2Asqrt%28-1%29%2Asqrt%285%29%29%2F%282%282%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%282%2B6i%2Asqrt%285%29%29%2F%282%282%29%29\" or \"x+=+%282-6i%2Asqrt%285%29%29%2F%282%282%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%282%281%2B3i%2Asqrt%285%29%29%29%2F%282%282%29%29\" or \"x+=+%282%281-3i%2Asqrt%285%29%29%29%2F%282%282%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281%2B3i%2Asqrt%285%29%29%2F2\" or \"x+=+%281-3i%2Asqrt%285%29%29%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "We end up with the roots that were given to us. So this confirms that \"2x%5E2-2x%2B23\" has the proper roots we were given. In other words, this confirms that the given roots lead to \"2x%5E2-2x%2B23\"\r
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\n" ); document.write( "\n" ); document.write( "Note: we could scale both sides of \"2x%5E2-2x%2B23+=+0\" by some nonzero number k, and get the same roots (since we can undo this and divide both sides by k). To keep things fairly simple, we'll make k = 1. We can consider \"2x%5E2-2x%2B23\" to be reduced as much as possible since the coefficients 2,-2,23 only have the common factor 1 between them. \r
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\n" ); document.write( "\n" ); document.write( "Answer: \"2x%5E2-2x%2B23+=+0\"
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