document.write( "Question 537580: I am having the hardest time with this. Solve by the method of your choice:\r
\n" ); document.write( "\n" ); document.write( "(1/x^2-3x+2)=(1/x+2)+(5/x^2-4)\r
\n" ); document.write( "\n" ); document.write( "I have tried this different ways and just cant figure it out.
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Algebra.Com's Answer #352920 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
Although you wrote (according to the rules of algebra) that you were given to solve:
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\n" ); document.write( "\"%281%2Fx%5E2-3x%2B2%29=%281%2Fx%2B2%29%2B%285%2Fx%5E2-4%29\"
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\n" ); document.write( "I'm guessing that you really were to solve the following:
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\n" ); document.write( "\"%281%2F%28x%5E2-3x%2B2%29%29=%281%2F%28x%2B2%29%29%2B%285%2F%28x%5E2-4%29%29\"
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\n" ); document.write( "If I'm wrong I apologize, and please ignore the following and re-submit your problem.
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\n" ); document.write( "Start with:
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\n" ); document.write( "\"%281%2F%28x%5E2-3x%2B2%29%29=%281%2F%28x%2B2%29%29%2B%285%2F%28x%5E2-4%29%29\"
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\n" ); document.write( "Where it is possible, factor the denominators. This will give you:
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\n" ); document.write( "Next, multiply both sides (all terms) by the three factors common to the denominator. Just multiply each of the terms on both sides by:
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\n" ); document.write( " \"%28x-1%29%28x%2B2%29%28x-2%29\"
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\n" ); document.write( "When you do that the equation becomes:
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\n" ); document.write( "Now in each term cancel any factors that are in both the numerator and the denominator:
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\n" ); document.write( "and you are left with:
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\n" ); document.write( "\"%28x%2B2%29+=+%28x-1%29%28x-2%29%2B+5%28x-1%29\"
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\n" ); document.write( "Multiply out the two terms on the right side to get:
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\n" ); document.write( "\"x%2B2+=+x%5E2-3x+%2B2%2B5x+-+5\"
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\n" ); document.write( "On the right side combine the -3x and +5x to get +2x. Also combine the +2 and the -5 to get -3. This reduces the equation to:
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\n" ); document.write( "\"x%2B2+=+x%5E2+%2B2x+-3\"
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\n" ); document.write( "Then get everything on one side of the equation by subtracting x + 2 from both sides. The equation then becomes:
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\n" ); document.write( "\"0+=+x%5E2+%2Bx+-5\"
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\n" ); document.write( "Transpose it to the standard quadratic form:
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\n" ); document.write( "\"x%5E2+%2Bx-5+=0\"
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\n" ); document.write( "Solve by using the quadratic formula:
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\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
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\n" ); document.write( "Recall that a is the multiplier of the \"x%5E2\". Therefore, a = 1. b is the multiplier of the x. Therefore b also = 1. And c is the constant. And so c = -5. Substituting these values into the quadratic formula results in:
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\n" ); document.write( "\"x+=+%28-%28%2B1%29+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-5%29+%29%29%2F%282%2A1%29+\"
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\n" ); document.write( "This becomes:
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\n" ); document.write( "\"x+=+%28-1+%2B-+sqrt%28+1%5E2%2B20+%29%29%2F2+\"
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\n" ); document.write( "So the two answers for x are:
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\n" ); document.write( "\"x+=+%28-1+%2B+sqrt%28+21+%29%29%2F2+\"
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\n" ); document.write( "and
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\n" ); document.write( "\"x+=+%28-1+-+sqrt%28+21+%29%29%2F2+\"
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\n" ); document.write( "Hope this helps you to clear up the places you had trouble with.
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