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 Post subject: Honors Algebra 2 Solving rational equations
PostPosted: Fri, 6 May 2011 02:56:13 UTC 
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Please help me solve this:

(1/x^2)-x^2
___________= 3/2

(1/x)+x


The teacher gave us a hint, but I still do not understand how to solve this problem.
(Hint: You may need to use factoring by division to solve this.)


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 Post subject: Re: Honors Algebra 2 Solving rational equations
PostPosted: Fri, 6 May 2011 02:58:04 UTC 
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Bloodprincess15 wrote:
Please help me solve this:

(1/x^2)-x^2
___________= 3/2

(1/x)+x


The teacher gave us a hint, but I still do not understand how to solve this problem.
(Hint: You may need to use factoring by division to solve this.)


Clear denominators then solve the resulting equation which is a polynomial.

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 Post subject:
PostPosted: Fri, 6 May 2011 03:04:09 UTC 
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what do you multiply the equation by to clear the denominators?


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PostPosted: Fri, 6 May 2011 03:05:38 UTC 
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Bloodprincess15 wrote:
what do you multiply the equation by to clear the denominators?


x.

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 Post subject:
PostPosted: Fri, 6 May 2011 03:07:08 UTC 
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x^2 actually.


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PostPosted: Fri, 6 May 2011 03:09:55 UTC 
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Matt wrote:
x^2 actually.


Ah yes, silly numerator fractions.

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PostPosted: Fri, 6 May 2011 03:12:22 UTC 
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So then would you get:

1-x^4
_____=3x^2/2x^2
x+x^3

Or did I multiply wrong?


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PostPosted: Fri, 6 May 2011 03:15:12 UTC 
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Bloodprincess15 wrote:
So then would you get:

1-x^4
_____=3x^2/2x^2
x+x^3

Or did I multiply wrong?


You didn't multiply the right side by x^2, you multiplied it by x^2/x^2, which is not the same thing.

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 Post subject:
PostPosted: Fri, 6 May 2011 03:21:13 UTC 
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so then it would be:

....=3x^2/2

Right?


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 Post subject:
PostPosted: Fri, 6 May 2011 03:35:52 UTC 
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Multiplying by x^2 on the left-hand side creates an equivalent fraction.
This has nothing to do with the right-hand side. The result is:

$\frac{1-x^4}{x+x^3}=\frac32


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 Post subject:
PostPosted: Fri, 6 May 2011 10:56:15 UTC 
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Princess, the "left side multiplication by x^2" means the numerator
AND denominator are multiplied; so that's really a multiplication by
x^2 / x^2 which equals 1, so no need to adjust right side...OK?

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 Post subject: Re: Honors Algebra 2 Solving rational equations
PostPosted: Fri, 6 May 2011 14:02:31 UTC 
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Hello, Bloodprincess15!

Another approach . . .


Quote:
Solve: .$\frac{\frac{1}{x^2} - x^2}{\frac{1}{x} + x} \:=\:\frac{3}{2}

Note that the numerator is a difference-of-squares.


$\text{Factor: }\;\frac{\left(\frac{1}{x} - x\right)\left(\frac{1}{x} + x\right)}{\frac{1}{x} + x} \;=\;\frac{3}{2}

Reduce: .$\frac{1}{x} - x \:=\:\frac{3}{2} . . (Note that: \frac{1}{x}+x \,\ne\,0 for real x.)


Multiply by 2x\!:\;\; 2 - 2x^2 \:=\:3x \quad\Rightarrow\quad 2x^2 + 3x - 2 \:=\:0


Therefore: .(x + 2)(2x-1) \:=\:0 \quad\Rightarrow\quad x \:=\:-2,\,\frac{1}{2}



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 Post subject:
PostPosted: Wed, 11 May 2011 19:05:43 UTC 
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I would have first cross multiplied, and then multiplied both sides by x^2.

Just another way to accomplish the same result.


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 Post subject:
PostPosted: Thu, 12 May 2011 07:37:29 UTC 
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Mr. Mathturo wrote:
I would have first cross multiplied, .........

YA!
Take that, TKHunny :shock:

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