anonymous
  • anonymous
(2+sqrt-16)(4-sqrt-9) Do the expression and express the answer in a+bi form.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Alex_Mattucci
  • Alex_Mattucci
could you draw this one out please?
anonymous
  • anonymous
\[(2+\sqrt{-16})(4-\sqrt{-9})\]
Alex_Mattucci
  • Alex_Mattucci
ok thank you

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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \left(2+\sqrt{-16}\right) \left(4-\sqrt{-9}\right) }\) you know that: \(\sqrt{-a^2}=\sqrt{a^2}\times\sqrt{-1}=ia\) And this way, it follows that: \(\bullet~~~~~~\sqrt{-9}=3i \) \(\bullet~~~~~~\sqrt{-16}=4i\)
SolomonZelman
  • SolomonZelman
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:
anonymous
  • anonymous
Before I continue, why did you put a 2 and 4i in front of the binomials?
SolomonZelman
  • SolomonZelman
Have you expanded expressions like: (A+B)(C+D) previously?
SolomonZelman
  • SolomonZelman
((You would multiply times A times C+D, and B times C+D. Then you add the results.))
anonymous
  • anonymous
yes but I don't understand how you did that
anonymous
  • anonymous
Oh!
SolomonZelman
  • SolomonZelman
Yes?
SolomonZelman
  • SolomonZelman
So can you continue from: \(\large\color{black}{ \displaystyle 2\left(4-3i\right)+4i\left(4-3i\right) }\) for me please?
anonymous
  • anonymous
Yes Im working it out now
anonymous
  • anonymous
I got a trinomial?
SolomonZelman
  • SolomonZelman
i is not a variable, it is a square root of -1. And it abides by the property: \(i^2=\left(\sqrt{-1}\right)^2=-1\)
SolomonZelman
  • SolomonZelman
So the term of i² is just a negative one.
SolomonZelman
  • SolomonZelman
Anyway, can you post what you have got?
anonymous
  • anonymous
when I work it out I get 8+22i which isn't the answer
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle 2\left(4-3i\right)+4i\left(4-3i\right) }\) \(\large\color{black}{ \displaystyle 2(4)+2(-3i)+4i(4)+4i(-3i) }\) \(\large\color{black}{ \displaystyle 8+(-6i)+(8i)+(-12i^2) }\)
anonymous
  • anonymous
4i(4) should be 16i why is it 8i?
anonymous
  • anonymous
This is where I got my answer wrong
SolomonZelman
  • SolomonZelman
Oh, sorry, my bad
SolomonZelman
  • SolomonZelman
yes, you were right.... \(\large\color{black}{ \displaystyle 8+(-6i)+(16i)+(-12i^2) }\) \(\large\color{black}{ \displaystyle 8+2i+(-12(-1)) }\) \(\large\color{black}{ \displaystyle 8+10i+(12) }\) \(\large\color{black}{ \displaystyle 20+10i }\)
anonymous
  • anonymous
Wow I made a really dumb mistake.. Thanks!
SolomonZelman
  • SolomonZelman
I did as well.

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