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anonymous

  • 5 years ago

How do you find all the real solutions to x^3-13x-12=0?

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  1. anonymous
    • 5 years ago
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    wait one moment

  2. anonymous
    • 5 years ago
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    \[x \approx4 ; x \approx-3 ; x \approx-1\]

  3. anonymous
    • 5 years ago
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    I was a bit complicated

  4. anonymous
    • 5 years ago
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    was rectified and that is the correct answer

  5. anonymous
    • 5 years ago
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    What about x^4-14x^2+45=0?

  6. anonymous
    • 5 years ago
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    wait one moment

  7. dumbcow
    • 5 years ago
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    factor (x^2 -5)(x^2 -9)=0 x^2 = 5 x^2 = 9

  8. anonymous
    • 5 years ago
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    x=3 x=-3 x=square root(5) x=-square root(5)

  9. anonymous
    • 5 years ago
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    \[\sqrt{5}\approx 2.23 ; -\sqrt{5}\approx - 2.23\]

  10. anonymous
    • 5 years ago
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    Could you still factor it if the 14x^2 was 14x^3(in the same problem)?

  11. anonymous
    • 5 years ago
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    wait one moment

  12. anonymous
    • 5 years ago
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    is very complicated

  13. anonymous
    • 5 years ago
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    wait one moment

  14. anonymous
    • 5 years ago
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    Don't worry about it then. I was just wondering if they could be only one power different or if they could also be two powers away from each other.

  15. anonymous
    • 5 years ago
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    solve this equation is very long process and I think I'm doing that I have out there in 10%

  16. anonymous
    • 5 years ago
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    expected and it started to let me finish lol

  17. anonymous
    • 5 years ago
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    Can you show me the steps to the x^4-7x^3=0 problem? i think I know how to get 7 but I don't know how you got sero as an answer.

  18. anonymous
    • 5 years ago
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    only by deduction. in looking for possible solutions when the right side of the expression as a result of 0 in this case is very easy to see that 7 is the only number other than 0 that satisfies this condition and so that the other three answers are 0 and which is a 4-degree equation must give 4 answers

  19. anonymous
    • 5 years ago
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    Okay, so you get x^3(x-7)=0 and solve the two pieces for the answers?

  20. anonymous
    • 5 years ago
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    ( x^3 ) (x-7)= 0 ??????

  21. anonymous
    • 5 years ago
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    I think that is how I was taught it. Then you get x=7 and x^3 =0

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