SOLUTION: I am having the hardest time with this. Solve by the method of your choice: (1/x^2-3x+2)=(1/x+2)+(5/x^2-4) I have tried this different ways and just cant figure it out.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I am having the hardest time with this. Solve by the method of your choice: (1/x^2-3x+2)=(1/x+2)+(5/x^2-4) I have tried this different ways and just cant figure it out.      Log On


   

Question 537580: I am having the hardest time with this. Solve by the method of your choice:
(1/x^2-3x+2)=(1/x+2)+(5/x^2-4)
I have tried this different ways and just cant figure it out.
Found 2 solutions by bucky, Tatiana_Stebko:
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Although you wrote (according to the rules of algebra) that you were given to solve:
.
%281%2Fx%5E2-3x%2B2%29=%281%2Fx%2B2%29%2B%285%2Fx%5E2-4%29
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I'm guessing that you really were to solve the following:
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%281%2F%28x%5E2-3x%2B2%29%29=%281%2F%28x%2B2%29%29%2B%285%2F%28x%5E2-4%29%29
.
If I'm wrong I apologize, and please ignore the following and re-submit your problem.
.
Start with:
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%281%2F%28x%5E2-3x%2B2%29%29=%281%2F%28x%2B2%29%29%2B%285%2F%28x%5E2-4%29%29
.
Where it is possible, factor the denominators. This will give you:
.

.
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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%28x-1%29%28x%2B2%29%28x-2%29
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When you do that the equation becomes:
.

.
Now in each term cancel any factors that are in both the numerator and the denominator:
.

.
and you are left with:
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%28x%2B2%29+=+%28x-1%29%28x-2%29%2B+5%28x-1%29
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Multiply out the two terms on the right side to get:
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x%2B2+=+x%5E2-3x+%2B2%2B5x+-+5
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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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x%2B2+=+x%5E2+%2B2x+-3
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Then get everything on one side of the equation by subtracting x + 2 from both sides. The equation then becomes:
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0+=+x%5E2+%2Bx+-5
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Transpose it to the standard quadratic form:
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x%5E2+%2Bx-5+=0
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Solve by using the quadratic formula:
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x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
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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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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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This becomes:
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x+=+%28-1+%2B-+sqrt%28+1%5E2%2B20+%29%29%2F2+
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So the two answers for x are:
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x+=+%28-1+%2B+sqrt%28+21+%29%29%2F2+
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and
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x+=+%28-1+-+sqrt%28+21+%29%29%2F2+
.
Hope this helps you to clear up the places you had trouble with.
.

Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
1%2F%28x%5E2-3x%2B2%29=1%2F%28x%2B2%29%2B5%2F%28x%5E2-4%29
1%2F%28%28x-2%29%28x-1%29%29-1%2F%28x%2B2%29-5%2F%28%28x-2%29%28x%2B2%29%29=0

%28x%2B2-%28x%5E2-3x%2B2%29-%285x-5%29%29%2F%28%28x-2%29%28x-1%29%28x%2B2%29%29=0
%28x%2B2-x%5E2%2B3x-2-5x%2B5%29%2F%28%28x-2%29%28x-1%29%28x%2B2%29%29=0
%28-x%5E2-x%2B5%29%2F%28%28x-2%29%28x-1%29%28x%2B2%29%29=0
The fraction is equal 0 when the numerator is equal 0 and denominator is not equal 0
-x%5E2-x%2B5=0 multiply by -1
x%5E2%2Bx-5=0
Use the quadratic formula x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=1, b=1, c=-5
x+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-5%29+%29%29%2F%282%2A1%29+
x+=+%28-1+%2B-+sqrt%28+21+%29%29%2F2+
x%5B1%5D+=+%28-1+-+sqrt%28+21+%29%29%2F2=-2.79+
x%5B2%5D+=+%28-1+%2Bsqrt%28+21+%29%29%2F2=1.79+