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anonymous

  • one year ago

Will give medal and become fan (accepting only correct answers)! Solve for x. (x^4 - 1) / (x^3) = 0

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  1. anonymous
    • one year ago
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    Please explain how you got the answer.

  2. zpupster
    • one year ago
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    multply both sides by x^3 factor (x^2-1)(x^2+1) and again (x^2-1) = (x+1)(x-1) set to zero x+1=0 x-1=0 What are your roots??? (x^2+1) can never be 0 when y=0 x is?? see graph

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  3. Nnesha
    • one year ago
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    okay so move x^3 to the right side as ~zpupster said multiply both sides by x^3

  4. anonymous
    • one year ago
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    Okay so then it would be \[x ^{3} \times (x ^{4} - 1) \div x ^{3} = 0\]

  5. anonymous
    • one year ago
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    But the x^3 can be cancelled so... the answer would be:\[x ^{4} - 1 = 0\]

  6. anonymous
    • one year ago
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    Right?

  7. Nnesha
    • one year ago
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    yep you need to multiply *both sides* so 0 times x^3 = 0 now apply difference of square rule

  8. Nnesha
    • one year ago
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    \[\huge\rm a^2-b^2 =(a+b)(a-b)\]

  9. anonymous
    • one year ago
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    \[(x ^{2} + 1)(x ^{2} - 1)\]

  10. anonymous
    • one year ago
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    \[(x ^{2}+1)(x+1)(x-1)\]

  11. anonymous
    • one year ago
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    Is that it?

  12. Nnesha
    • one year ago
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    yep right you would apply difference when there is negative sign so x^2-1 apply difference of square

  13. Nnesha
    • one year ago
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    oopps you already know

  14. Nnesha
    • one year ago
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    yes right set all 3 parentheses equal to zero solve for x

  15. anonymous
    • one year ago
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    How do you find the square root of -1?

  16. Nnesha
    • one year ago
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    well do you need for this question ?

  17. Nnesha
    • one year ago
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    ohh gotcha so you will get an imaginary solution

  18. anonymous
    • one year ago
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    Is it going to be \[\pm 1\]

  19. Nnesha
    • one year ago
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    \[\huge rm \sqrt{-1} = i\]

  20. anonymous
    • one year ago
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    Oh like that! So the answer is x = 1, -1, and i

  21. Nnesha
    • one year ago
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    \[\sqrt{-1}=i \]

  22. Nnesha
    • one year ago
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    yep

  23. anonymous
    • one year ago
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    Alright thanks again!

  24. Nnesha
    • one year ago
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    my pleasure :=)

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