Question

(2+sqrt-16)(4-sqrt-9)

Do the expression and express the answer in a+bi form.

Answer 1

could you draw this one out please?

Answer 2

\[(2+\sqrt{-16})(4-\sqrt{-9})\]

Answer 3

ok thank you

Answer 4

\(\large\color{black}{ \displaystyle \left(2+\sqrt{-16}\right) \left(4-\sqrt{-9}\right) }\)


you know that: \(\sqrt{-a^2}=\sqrt{a^2}\times\sqrt{-1}=ia\)

And this way, it follows that:
\(\bullet~~~~~~\sqrt{-9}=3i \)
\(\bullet~~~~~~\sqrt{-16}=4i\)

Answer 5

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:

Answer 6

Before I continue, why did you put a 2 and 4i in front of the binomials?

Answer 7

Have you expanded expressions like:
(A+B)(C+D)

previously?

Answer 8

((You would multiply times A times C+D, and B times C+D.
Then you add the results.))

Answer 9

yes but I don't understand how you did that

Answer 10

Oh!

Answer 11

Yes?

Answer 12

So can you continue from:

\(\large\color{black}{ \displaystyle 2\left(4-3i\right)+4i\left(4-3i\right) }\)

for me please?

Answer 13

Yes Im working it out now

Answer 14

I got a trinomial?

Answer 15

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

Answer 16

So the term of i² is just a negative one.

Answer 17

Anyway, can you post what you have got?

Answer 18

when I work it out I get 8+22i which isn't the answer

Answer 19

\(\large\color{black}{ \displaystyle 2\left(4-3i\right)+4i\left(4-3i\right) }\)
\(\large\color{black}{ \displaystyle 2(4)+2(-3i)+4i(4)+4i(-3i) }\)
\(\large\color{black}{ \displaystyle 8+(-6i)+(8i)+(-12i^2) }\)

Answer 20

4i(4) should be 16i why is it 8i?

Answer 21

This is where I got my answer wrong

Answer 22

Oh, sorry, my bad

Answer 23

yes, you were right....

\(\large\color{black}{ \displaystyle 8+(-6i)+(16i)+(-12i^2) }\)
\(\large\color{black}{ \displaystyle 8+2i+(-12(-1)) }\)
\(\large\color{black}{ \displaystyle 8+10i+(12) }\)
\(\large\color{black}{ \displaystyle 20+10i }\)

Answer 24

Wow I made a really dumb mistake.. Thanks!

Answer 25

I did as well.