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anonymous

  • one year ago

I need help finding the restriction on the quotient of a rational expression!

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  1. anonymous
    • one year ago
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    \[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{ 6x+12 }{ x+2 }\]

  2. terenzreignz
    • one year ago
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    Shame I don't know what "restriction on the quotient of a rational expression" means -_- However, you CAN simplify that divisor into just a simple number...

  3. anonymous
    • one year ago
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    Ive already multiplied by the reciprocal and then taken out the GCF of the numerator and the denominator and then factored the denominator leaving me with \[\frac{ 5(x^3-7x-6) }{ 6(x^2+9)(x+2) }\]

  4. anonymous
    • one year ago
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    My options are \[x \neq-2\] \[x \neq9\] \[x \neq3\] \[x \neq-3\] and i have to pick the one that is NOT a restriction

  5. terenzreignz
    • one year ago
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    Oh? It's easier than I thought. ^^ There's actually no need for any simplification. Just say, try the first choice: x = -2 Substitute it for the x's in the division and see if it results in anything you might call "illegal"

  6. terenzreignz
    • one year ago
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    And just so you know "illegal" is when you get a zero in the denominator, or a negative number inside a square root sign (irrelevant, since you don't have square roots here)

  7. anonymous
    • one year ago
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    Thats what i thought, but the only one that i can see IS a restriction is -2

  8. terenzreignz
    • one year ago
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    Good, so the first choice isn't the answer. What about the second? Does replacing the x's with 9's result in any...problems?

  9. anonymous
    • one year ago
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    nope

  10. terenzreignz
    • one year ago
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    So doesn't that mean the second one is the answer? haha Try the third option. Does substituting 3 for x cause problems?

  11. anonymous
    • one year ago
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    I thought it would be 9 too but then i tried 3 and that doesn't cause any 0's either

  12. terenzreignz
    • one year ago
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    Are you sure? ;) Look closely at this part: \[\large \frac{ 5x^2-10x-15 }{ \boxed{\color{red}{x^2-9}} } \div \frac{ 6x+12 }{ x+2 }\]

  13. anonymous
    • one year ago
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    Im supposed to substitute in the original equation???? i thought i was supposed to simplify first....

  14. terenzreignz
    • one year ago
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    You could simplify, but in these cases, original expression is king. And I'll tell you why: If x was, say, -2, then the original equation would have been illegal in the first place, as, therefore, would any simplification of it. So the restrictions have to be binding on the original equation. To be more specific, we could say, generally, I could factor out the 6 here: \[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{ \color{blue}{6(x+2)} }{ x+2 }\] and then cancel out: \[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{ 6\cancel{(x+2)} }{\cancel{ x+2 }}\] But this would have been illegal if x was -2. So yeah, use the original expression when considering restrictions ^^

  15. anonymous
    • one year ago
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    THANK YOU SO MUCH OH MY GOD YOU HAVE NO IDEA HOW MUCH TROUBLE THIS SAVED ME

  16. terenzreignz
    • one year ago
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    No problem ^^ I trust you have no doubts as to your answer now? :D

  17. anonymous
    • one year ago
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    Nope (; i got it! Thanks so much!!!

  18. terenzreignz
    • one year ago
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    One more thing... and this doesn't really affect THIS particular problem, but I just thought you'd know... since we ARE dividing rational expressions, any value of x that causes the NUMERATOR of the DIVISOR to be equal to zero must also be illegal. Specifically, this part: \[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{\boxed{\color{green}{ 6x+12}} }{ x+2 }\] That's because that also results in a division by zero when you consider the whole thing. Get it? Got it? Good ^^ Good job, by the way :D

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