document.write( "Question 210323This question is from textbook COLLEGE ALGEBRA
\n" ); document.write( ": sorry I already ask this question but i cannot find it again
\n" ); document.write( " (x+ 1/x)^2 -2(x+ 1/x)+1=0 this part of equation in quadratic form
\n" ); document.write( "I already got an answer x^4-2x^3+3x^2-2x+1=0 But i got stuck here i dont know what to do next \r
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Algebra.Com's Answer #158931 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"%28x%2B+1%2Fx%29%5E2+-2%28x%2B+1%2Fx%29%2B1=0\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "Take note that the term \"x%2B1%2Fx\" is being repeated here. So one way to simplify this equation (for now) is to do a simple substitution\r
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\n" ); document.write( "\n" ); document.write( "So let \"z=x%2B1%2Fx\". So this means that \"z%5E2=%28x%2B1%2Fx%29%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "\"z%5E2-2z%2B1=0\" Replace each instance of \"x%2B1%2Fx\" with 'z'\r
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\n" ); document.write( "\n" ); document.write( "Now we have a much simpler equation to solve for.\r
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\n" ); document.write( "\n" ); document.write( "Notice that the quadratic \"z%5E2-2z%2B1\" is in the form of \"Az%5E2%2BBz%2BC\" where \"A=1\", \"B=-2\", and \"C=1\"\r
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\n" ); document.write( "\n" ); document.write( "Let's use the quadratic formula to solve for \"z\":\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%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( "\"z+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%281%29%281%29+%29%29%2F%282%281%29%29\" Plug in \"A=1\", \"B=-2\", and \"C=1\"\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%281%29%281%29+%29%29%2F%282%281%29%29\" Negate \"-2\" to get \"2\". \r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%282+%2B-+sqrt%28+4-4%281%29%281%29+%29%29%2F%282%281%29%29\" Square \"-2\" to get \"4\". \r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%282+%2B-+sqrt%28+4-4+%29%29%2F%282%281%29%29\" Multiply \"4%281%29%281%29\" to get \"4\"\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%282+%2B-+sqrt%28+0+%29%29%2F%282%281%29%29\" Subtract \"4\" from \"4\" to get \"0\"\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%282+%2B-+sqrt%28+0+%29%29%2F%282%29\" Multiply \"2\" and \"1\" to get \"2\". \r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%282+%2B-+0%29%2F%282%29\" Take the square root of \"0\" to get \"0\". \r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%282+%2B+0%29%2F%282%29\" or \"z+=+%282+-+0%29%2F%282%29\" Break up the expression. \r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%282%29%2F%282%29\" or \"z+=++%282%29%2F%282%29\" Combine like terms. \r
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\n" ); document.write( "\n" ); document.write( "\"z+=+1\" or \"z+=+1\" Simplify. \r
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\n" ); document.write( "\n" ); document.write( "So the only solution in terms of 'z' is \"z+=+1\" \r
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\n" ); document.write( "\n" ); document.write( "However, we want to solve for 'x' (not 'z'). But do recall that we let \"z=x%2B1%2Fx\"\r
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\n" ); document.write( "\n" ); document.write( "\"z=x%2B1%2Fx\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"1=x%2B1%2Fx\" Plug in the solution (in terms of 'z') \"z=1\"\r
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\n" ); document.write( "\n" ); document.write( "\"x=x%5E2%2B1\" Multiply EVERY term by the LCD 'x' to clear out the fraction.\r
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\n" ); document.write( "\n" ); document.write( "\"0=x%5E2-x%2B1\" Subtract 'x' from both sides.\r
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\n" ); document.write( "\n" ); document.write( "Notice that the quadratic \"x%5E2-x%2B1\" is in the form of \"Ax%5E2%2BBx%2BC\" where \"A=1\", \"B=-1\", and \"C=1\"\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%281%29+%29%29%2F%282%281%29%29\" Plug in \"A=1\", \"B=-1\", and \"C=1\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+%28-1%29%5E2-4%281%29%281%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%281%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+1-4+%29%29%2F%282%281%29%29\" Multiply \"4%281%29%281%29\" to get \"4\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+-3+%29%29%2F%282%281%29%29\" Subtract \"4\" from \"1\" to get \"-3\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281+%2B-+sqrt%28+-3+%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+%2B-+i%2Asqrt%283%29%29%2F%282%29\" Simplify the square root \r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281%2Bi%2Asqrt%283%29%29%2F%282%29\" or \"x+=+%281-i%2Asqrt%283%29%29%2F%282%29\" Break up the expression. \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the solutions (in terms of x) are \"x+=+%281%2Bi%2Asqrt%283%29%29%2F%282%29\" or \"x+=+%281-i%2Asqrt%283%29%29%2F%282%29\"\r
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\n" ); document.write( "\n" ); document.write( "Note: since the degree of \"%28x%2B+1%2Fx%29%5E2+-2%28x%2B+1%2Fx%29%2B1\" turns out to be 4, this means that there are really 4 solutions. So what happened? The twist is that each solution is counted twice. You'll most likely hear that these solutions are of multiplicity 2.
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