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OpenStudy (anonymous):
Need to know how to solve a rational equation
12 years ago
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OpenStudy (anonymous):
The following equation is one that I made up so I can solve the problem I have to do on my own.
12 years ago
OpenStudy (anonymous):
|dw:1407178562891:dw| this is the equation I made
12 years ago
OpenStudy (anonymous):
|dw:1407178841444:dw|
12 years ago
OpenStudy (anonymous):
I don't know why it came out like that
12 years ago
OpenStudy (agreene):
I would start like this:
\[(x^3-8x)^{-1}=\frac{x}x+\frac8x-6\]
12 years ago
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OpenStudy (anonymous):
okay hold on let me put the other equation in that form @agreene
12 years ago
OpenStudy (anonymous):
Where did the ^-1 outside the first parenthesis come from? @agreene
12 years ago
OpenStudy (agreene):
its shorthand for the division
12 years ago
OpenStudy (agreene):
\[\frac1x=x^{-1}\]
12 years ago
OpenStudy (anonymous):
so the ^-1 is coming from the 1 in the first part of the equation?
12 years ago
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OpenStudy (agreene):
yes
12 years ago
OpenStudy (anonymous):
okay so I put my equation in that form whats next?
12 years ago
OpenStudy (anonymous):
you there? @agreene
12 years ago
OpenStudy (anonymous):
what do I doo next?
12 years ago
OpenStudy (agreene):
lol im trying to remember.. im a bit rusty on the algebra here
12 years ago
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OpenStudy (anonymous):
haha its okay
12 years ago
OpenStudy (agreene):
\[\frac1{x^3-8x}=1+\frac8x-6\]
\[\frac1{x^3-8x}=\frac8x-5\]
\[\frac1{x^3-8x}-\frac8x=-5\]
12 years ago
OpenStudy (aum):
The least common denominator is (x^3-8x).
Multiply throughout by (x^3-8x).
You will get a fourth degree polynomial which cannot be easily solved unless you use a graphic calculator.
Note that x cannot be 0 or \(\pm\sqrt{8}\) because that would make the denominator zero.
12 years ago
OpenStudy (anonymous):
My equation was
12 years ago
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