Question

can someone explain this to me please

Answer 1

\[\sqrt{-72}\]

Answer 2

@mathmate

Answer 3

@mudslinger
What do you know about i (in mathematics)?

Answer 4

it repeats every 4 roups and it equals\[\sqrt{-1}\]

Answer 5

Exactly!
So
\(i=\sqrt{-1}\)
\(i^2=\sqrt{-1}^2=-1\)
\(i^3=\sqrt{-1}^3=-i\)
\(i^4=\sqrt{-1}^4=+1\)
Agree?

Answer 6

is there an easy way to remember that

Answer 7

Just do the multiplication step by step, no need to remember.
In my books, apart from the multiplication table, there is nothing to memorize in math.
Some people can even multiply without the multiplication table.

Answer 8

ok i understand that

Answer 9

Next, can you tell me:
\(\sqrt{a}\times\sqrt{b}=?\)

Answer 10

no its only mentioned i so far

Answer 11

@mudslinger
Remember, in math, we have to use everything we have learned before. This is called cumulative knowledge. What you have learned, last month, last year, or when you were in grade school, are all necessary to solve a problem today.
Imagine you cannot multiply 4\(\times\)5 today, you'd be in big trouble!

Answer 12

ok

Answer 13

Now,
\(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\)
which means that
\(\sqrt{-72}=\sqrt{-1}\times\sqrt{72}\)
ok so far?

Answer 14

ok

Answer 15

\(\sqrt{-1}\times\sqrt{72}=~?~\times\sqrt{72}\)
can you do that?

Answer 16

no

Answer 17

What does \(\sqrt{-1}\) equal?
Look back earlier in the thread if you need to.

Answer 18

1

Answer 19

Hm.. you want to try again?

Answer 20

oh 1i

Answer 21

good!
So now
\(\sqrt{-72}=\sqrt{-1}\times\sqrt{72}=\sqrt{72} i\)
so far so good?

Answer 22

yeah i think i get it

Answer 23

But it's not done yet, we need to do the same to split up the real number 72.
Can you give it a try, hint: 72=8\(\times\)9=\(2^3\times 3^2\)

Answer 24

ok i get the 8x9=72 and the 2 cubed and 3 squared but whet do these two have to do with it

Answer 25

Here's an example:
\(\sqrt{36}=\sqrt{4}\times\sqrt{9}=2\times 3\)

Answer 26

ok

Answer 27

\(\sqrt{-72}=\sqrt{-1}\times\sqrt{72}=\sqrt{72} i=\sqrt{2}\times \sqrt{4}\times\sqrt{9}~ i=?\)

Answer 28

\[\sqrt{-72}\]

Answer 29

Well, that's our starting point, we are progressing left to right.
Can you work from the expression just before the question mark (?), reviewing the example I just gave you a moment ago, as you do that?

Answer 30

what you mean

Answer 31

Try to replace the parts in red with something else.
\(\sqrt{-72}=\sqrt{-1}\times\sqrt{72}=\sqrt{72} i=\sqrt{2}\times \color{red}{\sqrt{4}}\times\color{red}{\sqrt{9}}~ i=?\)

Answer 32

6

Answer 33

You're getting close,
but you need to \(replace\), meaning the rest of the expression must be there too!

\(\sqrt{2}\times \color{red}{\sqrt{4}}\times\color{red}{\sqrt{9}}~ i=?\)

Answer 34

\[\sqrt{6} \sqrt{6}\]

Answer 35

\(\sqrt{2}\times \color{red}{\sqrt{4}}\times\color{red}{\sqrt{9}}~ i\)
\(\sqrt{2}\times 2\times3~ i=?\)
I will let you finish the final answer.
I have to go now, and I'll check your answer when I come back, or someone else would be nice enough to check your answer.

Answer 36

ill tag you in it

Answer 37

@mudslinger
Thought this might be helpful for you.
http://www.themathpage.com/alg/multiply-radicals.htm