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anonymous
 one year ago
(2+sqrt16)(4sqrt9)
Do the expression and express the answer in a+bi form.
anonymous
 one year ago
(2+sqrt16)(4sqrt9) Do the expression and express the answer in a+bi form.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could you draw this one out please?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(2+\sqrt{16})(4\sqrt{9})\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle \left(2+\sqrt{16}\right) \left(4\sqrt{9}\right) }\) 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
 one year ago
Best ResponseYou've already chosen the best response.0So, you can rewrite your expression is: \(\large\color{black}{ \displaystyle \left(2+4i\right) \left(43i\right) }\) then, expand the expression: \(\large\color{black}{ \displaystyle 2\left(43i\right)+4i\left(43i\right) }\) continue expanding, and tell me what you get:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Before I continue, why did you put a 2 and 4i in front of the binomials?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Have you expanded expressions like: (A+B)(C+D) previously?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0((You would multiply times A times C+D, and B times C+D. Then you add the results.))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but I don't understand how you did that

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0So can you continue from: \(\large\color{black}{ \displaystyle 2\left(43i\right)+4i\left(43i\right) }\) for me please?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes Im working it out now

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0So the term of i² is just a negative one.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Anyway, can you post what you have got?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when I work it out I get 8+22i which isn't the answer

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle 2\left(43i\right)+4i\left(43i\right) }\) \(\large\color{black}{ \displaystyle 2(4)+2(3i)+4i(4)+4i(3i) }\) \(\large\color{black}{ \displaystyle 8+(6i)+(8i)+(12i^2) }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04i(4) should be 16i why is it 8i?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is where I got my answer wrong

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Oh, sorry, my bad

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0Wow I made a really dumb mistake.. Thanks!
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