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

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

Mathematics
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could you draw this one out please?
\[(2+\sqrt{-16})(4-\sqrt{-9})\]
ok thank you

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

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