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anonymous

  • one year ago

Solve (z + 6)2 = 5. A. B. C. D.

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  1. anonymous
    • one year ago
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    Might help: http://www.mathpapa.com/algebra-calculator.html also shows you step by step

  2. anonymous
    • one year ago
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    |dw:1432838616284:dw|

  3. anonymous
    • one year ago
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    i get confused after this?@KendrickLamar2014

  4. anonymous
    • one year ago
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    the answer choices didnt go through, i will wrte them out through the draw tool! @KendrickLamar2014

  5. anonymous
    • one year ago
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    |dw:1432838835991:dw| @KendrickLamar2014

  6. anonymous
    • one year ago
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    @KendrickLamar2014

  7. KendrickLamar2014
    • one year ago
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    My guess would be choice C, not 100% sure tho.

  8. KendrickLamar2014
    • one year ago
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    @Michele_Laino

  9. anonymous
    • one year ago
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    im in between B and c?

  10. KendrickLamar2014
    • one year ago
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    @Luigi0210 @kropot72

  11. anonymous
    • one year ago
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    help!!! @KendrickLamar2014 @ssssss @Vocaloid @AriPotta

  12. Vocaloid
    • one year ago
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    OH MY GOODNESS, ok, I didn't realize that the 2 was an exponent!

  13. Vocaloid
    • one year ago
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    that changes everything! (z + 6)^2 = 5

  14. Vocaloid
    • one year ago
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    take the square root of each side subtract 6 from each side

  15. Luigi0210
    • one year ago
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    \(\Large (z+6)^2=5\) \(\Large \sqrt{(z+6)^2}=\sqrt{5} \) \(\Large z+6=\pm \sqrt{5} \) \(\Large z+6-6=\pm\sqrt{5}-6\) \(\Large \color{green}{z=\pm\sqrt{5}-6}\)

  16. KendrickLamar2014
    • one year ago
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    Oh the 2 is an exponent. No wonder I could not find the answer :P

  17. anonymous
    • one year ago
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    oh!

  18. Luigi0210
    • one year ago
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    Good luck \(\Huge \color{red}{\star^{\star}}\)

  19. KendrickLamar2014
    • one year ago
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    So the answer is: ______

  20. anonymous
    • one year ago
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    @Luigi0210 thats not an answer choice but whould it be c? its the closest one?

  21. Vocaloid
    • one year ago
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    hint: A + B = B + A

  22. Michele_Laino
    • one year ago
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    is z a complex number?

  23. Vocaloid
    • one year ago
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    when you add two terms together you can switch the order

  24. anonymous
    • one year ago
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    okay so my best assumption would be b or c?

  25. anonymous
    • one year ago
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    yes it is what it will equal up to!:) @Michele_Laino

  26. Luigi0210
    • one year ago
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    It's magic! \(\Large \pm \sqrt{5}-6 = \color{red}{ -6\pm \sqrt{5}}=\color{green}{-6+\sqrt{5} ~or~ -6-\sqrt{5}}\)

  27. Michele_Laino
    • one year ago
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    if z is a complex number, then your equation, can be rewritten as below: \[\Large {\left( {x + 6 + iy} \right)^2} = 5\] wher i is such that: \[\Large {i^2} = - 1\] and I have used: \[\Large z = x + iy\]

  28. Michele_Laino
    • one year ago
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    so we are led to this subsequent algebraic system: \[\Large \left\{ {\begin{array}{*{20}{c}} {\left( {x + 6} \right)y = 0} \\ {{{\left( {x + 6} \right)}^2} - {y^2} = 5} \end{array}} \right.\] whose acceptable solutions are: \[\Large \left\{ \begin{gathered} {x_1} = - 6 + \sqrt 5 \hfill \\ {y_1} = 0 \hfill \\ \end{gathered} \right.,\quad \left\{ {\begin{array}{*{20}{c}} {{x_2} = - 6 - \sqrt 5 } \\ {{y_2} = 0} \end{array}} \right.\]

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