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Jamierox4ev3r

  • one year ago

More review! Help appreciated

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  1. Jamierox4ev3r
    • one year ago
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    4e. Factor the following expression: \[3x ^{\frac{ 3 }{ 2 }} -9x ^{\frac{ 1 }{ 2 }} +6x ^{-\frac{ 1 }{ 2 }}\]

  2. Jamierox4ev3r
    • one year ago
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    I'm pretty lost on this, having to factor with the rational fractions is throwing me into a bit of a loop.

  3. Jamierox4ev3r
    • one year ago
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    Does anyone have a simple method of doing this by hand? Or do you recommend grabbing a calculator to look for x-intercepts so I can factor that way?

  4. anonymous
    • one year ago
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    hold on i gotta write it :)

  5. Jamierox4ev3r
    • one year ago
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    fair enough, thank you @PinkiePug

  6. anonymous
    • one year ago
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    \[3(x \frac{ 1 }{ 2 } - 3x \frac{ 1 }{ 2 }+2 x \frac{ 1 }{ 2 }\] this is what i got

  7. Astrophysics
    • one year ago
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    \[x^{1/2} \implies \sqrt{x}\] \[x^{-1/2} \implies \frac{ 1 }{ \sqrt{x} }\] This may be easier to visualize, so we can write it as \[3x^{3/2}-9\sqrt{x}+6\frac{ 1 }{ \sqrt{x} }\]

  8. Astrophysics
    • one year ago
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    Then we can find the common denominator \[\sqrt{x}\]

  9. Astrophysics
    • one year ago
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    Your numerator should simplify into something better which you can factor

  10. Jamierox4ev3r
    • one year ago
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    Wait. so are you suggesting that we have to get rid of the negative exponents in order to successfully factor this?

  11. phi
    • one year ago
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    I would factor out 3x^-1/2 \[ 3x ^{\frac{ 3 }{ 2 }} -9x ^{\frac{ 1 }{ 2 }} +6x ^{-\frac{ 1 }{ 2 }} \\ 3x ^{-\frac{ 1 }{ 2 }}\left( x^2 -3x +2\right) \]

  12. phi
    • one year ago
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    now you can factor the quadratic

  13. Astrophysics
    • one year ago
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    Yeah that works to haha :P

  14. Astrophysics
    • one year ago
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    \[\frac{ 3x^2-9x+6 }{ \sqrt{x} }\] is what you get from my suggestion

  15. Jamierox4ev3r
    • one year ago
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    oh. so i can factor \(x^{2}-3x+2\), since that's the quadratic. Oh wow that's much simpler. So the final answer would be: \(3x^{-\Large\frac{1}{2}}\)(x-1)(x-2)

  16. phi
    • one year ago
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    yes, or put sqr(x) in the denominator it is still ugly. but it's factored

  17. Jamierox4ev3r
    • one year ago
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    so just so I know, for future problems like this, is it always acceptable to take out a value so that you're left with a quadratic that you can easily factor?

  18. phi
    • one year ago
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    it is a common trick. no guarantees, but something to try

  19. Jamierox4ev3r
    • one year ago
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    Fair enough. I'll keep that in mind. Thank you very much kind sir

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