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anonymous

  • one year ago

(2+sqrt-16)(4-sqrt-9) Do the expression and express the answer in a+bi form.

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  1. Alex_Mattucci
    • one year ago
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    could you draw this one out please?

  2. anonymous
    • one year ago
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    \[(2+\sqrt{-16})(4-\sqrt{-9})\]

  3. Alex_Mattucci
    • one year ago
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    ok thank you

  4. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \left(2+\sqrt{-16}\right) \left(4-\sqrt{-9}\right) }\) you know that: \(\sqrt{-a^2}=\sqrt{a^2}\times\sqrt{-1}=ia\) And this way, it follows that: \(\bullet~~~~~~\sqrt{-9}=3i \) \(\bullet~~~~~~\sqrt{-16}=4i\)

  5. SolomonZelman
    • one year ago
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    So, you can re-write your expression is: \(\large\color{black}{ \displaystyle \left(2+4i\right) \left(4-3i\right) }\) then, expand the expression: \(\large\color{black}{ \displaystyle 2\left(4-3i\right)+4i\left(4-3i\right) }\) continue expanding, and tell me what you get:

  6. anonymous
    • one year ago
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    Before I continue, why did you put a 2 and 4i in front of the binomials?

  7. SolomonZelman
    • one year ago
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    Have you expanded expressions like: (A+B)(C+D) previously?

  8. SolomonZelman
    • one year ago
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    ((You would multiply times A times C+D, and B times C+D. Then you add the results.))

  9. anonymous
    • one year ago
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    yes but I don't understand how you did that

  10. anonymous
    • one year ago
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    Oh!

  11. SolomonZelman
    • one year ago
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    Yes?

  12. SolomonZelman
    • one year ago
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    So can you continue from: \(\large\color{black}{ \displaystyle 2\left(4-3i\right)+4i\left(4-3i\right) }\) for me please?

  13. anonymous
    • one year ago
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    Yes Im working it out now

  14. anonymous
    • one year ago
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    I got a trinomial?

  15. SolomonZelman
    • one year ago
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    i is not a variable, it is a square root of -1. And it abides by the property: \(i^2=\left(\sqrt{-1}\right)^2=-1\)

  16. SolomonZelman
    • one year ago
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    So the term of i² is just a negative one.

  17. SolomonZelman
    • one year ago
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    Anyway, can you post what you have got?

  18. anonymous
    • one year ago
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    when I work it out I get 8+22i which isn't the answer

  19. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle 2\left(4-3i\right)+4i\left(4-3i\right) }\) \(\large\color{black}{ \displaystyle 2(4)+2(-3i)+4i(4)+4i(-3i) }\) \(\large\color{black}{ \displaystyle 8+(-6i)+(8i)+(-12i^2) }\)

  20. anonymous
    • one year ago
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    4i(4) should be 16i why is it 8i?

  21. anonymous
    • one year ago
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    This is where I got my answer wrong

  22. SolomonZelman
    • one year ago
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    Oh, sorry, my bad

  23. SolomonZelman
    • one year ago
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    yes, you were right.... \(\large\color{black}{ \displaystyle 8+(-6i)+(16i)+(-12i^2) }\) \(\large\color{black}{ \displaystyle 8+2i+(-12(-1)) }\) \(\large\color{black}{ \displaystyle 8+10i+(12) }\) \(\large\color{black}{ \displaystyle 20+10i }\)

  24. anonymous
    • one year ago
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    Wow I made a really dumb mistake.. Thanks!

  25. SolomonZelman
    • one year ago
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    I did as well.

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