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anonymous

  • 4 years ago

how do you solve 3t^6-48t^2=0

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  1. Mertsj
    • 4 years ago
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    Factor out 3t^2

  2. Mertsj
    • 4 years ago
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    3t^2(t^4-16)

  3. cwrw238
    • 4 years ago
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    factor 3t^2(t^4 - 16) = 0 t = 0 , 2 , -2

  4. Mertsj
    • 4 years ago
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    Now you have the difference of two squares. Factor that.

  5. anonymous
    • 4 years ago
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    How do you factor that out? I understand how you get that but there should be six answers

  6. anonymous
    • 4 years ago
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    t = 0 , 2 , -2, 2i,-2i

  7. Mertsj
    • 4 years ago
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    \[t^4-16=(t^2-4)(t^2+4)=(t-2)(t+2)(t^2+4)\]

  8. anonymous
    • 4 years ago
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    what happened to the 3t^2?

  9. anonymous
    • 4 years ago
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    x=0 double root

  10. Mertsj
    • 4 years ago
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    I left it off because I thought you already recognized that it is a factor.

  11. anonymous
    • 4 years ago
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    so how do you get the six answers by factoring is what im confused about

  12. Mertsj
    • 4 years ago
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    \[3t^2(t-2)(t+2)(t^2+4)\]

  13. Mertsj
    • 4 years ago
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    That is the completely factored form.

  14. anonymous
    • 4 years ago
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    \[x_1=0,x_2=0,x_3=2,x_4=-2,x_5=2i,x_6=-2i\]

  15. anonymous
    • 4 years ago
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    how do you get the six answers though. I know there should be six because the degree is six

  16. anonymous
    • 4 years ago
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    there are 6 answers already!

  17. anonymous
    • 4 years ago
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    How didyou get them though? I see and understand there are six but i dont know HOW you got them

  18. anonymous
    • 4 years ago
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    \[3t^6-48t^2=0\] \[3t^2(t^4-16)=0\] \[3t^2(t-2)(t+2)(t^2+4)=0\]

  19. Mertsj
    • 4 years ago
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    If the product of factors is 0 then one of the factors has to be 0. so set each factor to 0 and solve

  20. phi
    • 4 years ago
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    not sure how you are confused. Do you know how to find the *2* roots of \[ t^2= 0\] t= ±0 two solutions, 0 and 0

  21. anonymous
    • 4 years ago
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    \[3t^2=0\] \[t+2=0\] \[t-2=0\] \[t^2+4=0\]

  22. anonymous
    • 4 years ago
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    if a*b=0 then a=0 or b=0

  23. anonymous
    • 4 years ago
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    Okay, now I understand, thankyou so much

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