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anonymous

  • 5 years ago

what is the square root of x^4 + 7?

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    okay so 4+7 is 11 and the square root of 11 is? what are 2 of the same #s that divide into 11?like the square root of 36 is 6 bcuz 6 times 6 is 36.

  3. anonymous
    • 5 years ago
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    (x^4 + 7)^1/2

  4. radar
    • 5 years ago
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    \[\sqrt{x ^{4}+7}\] or\[\sqrt{x ^{4+7}}\]??

  5. anonymous
    • 5 years ago
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    na I need a specific answer

  6. anonymous
    • 5 years ago
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    it is apparently factorable

  7. anonymous
    • 5 years ago
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    There is no good way to simplify that problem. It is simply \[\sqrt(x^4+7)\] Owlfred, your reply was useless. Helpneededlady, 11 is prime, and that was not this persons question anyway. Jman, is there more to the problem?

  8. anonymous
    • 5 years ago
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    no

  9. anonymous
    • 5 years ago
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    thats it I have to simplify it further

  10. anonymous
    • 5 years ago
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    Find f(√x + 2) if f(x) = x4 + 5

  11. anonymous
    • 5 years ago
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    that is the problem

  12. anonymous
    • 5 years ago
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    only it is x to the fourth

  13. anonymous
    • 5 years ago
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    Well then you already gave us the solution to your problem, and it cannot be simplified further.

  14. anonymous
    • 5 years ago
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    no it has to be simplified

  15. anonymous
    • 5 years ago
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    I did notice some calculation errors, that could be your problem. The answer ends up being \[(\sqrt(x+2))^4+5\]. At this point you can simplify and get a different answer.

  16. anonymous
    • 5 years ago
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    f(√x + 2) if f(x) = x4 + 5 (√x + 2)^4 + 5 = (x+2)^2 +5 = x^2 + 2x +9

  17. anonymous
    • 5 years ago
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    i dont think we r allowed to do that suzi

  18. anonymous
    • 5 years ago
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    why? f(√x + 2) means x = √x + 2 then plug into equation

  19. anonymous
    • 5 years ago
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    suzi, (sqrtx + 2)^4 is not (x + 2)^2

  20. anonymous
    • 5 years ago
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    Guyc is correct, my bad I didn't notice the calculation errors, but her general procedure was correct.

  21. anonymous
    • 5 years ago
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    √x + 2 = (x+2)^1/2 (√x + 2)^4 = ((x+2)^1/2)^4 (x+2)^ 1/2 * 4 (x+2)^2

  22. anonymous
    • 5 years ago
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    it was wrong

  23. anonymous
    • 5 years ago
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    ok up to u, that was my answer and thank you.

  24. anonymous
    • 5 years ago
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    The difficulty we're having is calculating (sqrt x + 2) ^4. Multiplying by hand by brute force I get: x^2 + 8x^(3/2) + 24x + 32x^(1/2) + 16 which is horrible

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