Question

Can I get some help please?

What is the product of three square root of eight times four square root of three? Simplify your answer.

Twenty four square root six
Twelve square root twelve
Twelve square root eleven
Seventy two square root four

Answer 1

do you mean
\[ 3 \sqrt{8} \cdot 4 \sqrt{3} \]

Answer 2

\[\huge\bf 3\sqrt{8} * 4\sqrt{3}?\] does it looks like tihs?

Answer 3

yes

Answer 4

we combine 3 with 4 and sqrt8 with sqrt3:

\[\large\rm so~we~have~~(3*4)*\sqrt{8*3}\]

Answer 5

ok

Answer 6

simplify this

Answer 7

29.39

Answer 8

when you do "square root" problems, it is usually helpful to factor the numbers inside the square root
for example, write 8 as 2*2*2
and 8*3 as 2*2*2*3
and then look for "pairs" of numbers to "pull out"

Answer 9

no, write this in sqrt form. do not convert it into decimal

Answer 10

how?

Answer 11

well what is 8*3?

Answer 12

24

Answer 13

@12wishs so there is 3sqrt8
than you know that sqrt8 = sqrt(4*2)
but you know that srt(4) = 2
so than how can you rewriting 3sqrt8 = ?

Answer 14

2^3?

Answer 15

sqrt8 = sqrt(2^2 *2) = ?

Answer 16

8

Answer 17

for example sqrt12 = sqrt(4*3) = sqrt(2^2 *3) = 2sqrt3
us this in case of your exercise please

Answer 18

There are several ways in which you could simplify "\[3 \sqrt{8} \cdot 4 \sqrt{3}\]

Answer 19

...and at least two have been mentioned by others.

Method 1: You can combine the two radicals Sqrt(8) and Sqrt(3) into a single radical:\[\sqrt{8}\sqrt{3}\sqrt{24}\]

Answer 20

That's = to Sqrt(24), sorry.

Answer 21

@mathmale there you have missed a sign of =

Answer 22

yeah

Answer 23

Next, in this same example, we can factor 24 into one factor that IS a perfect square and one factor which is not. Could you do that, please?

Answer 24

4 is a perfect square and can be considered the square root of 24 and 3 can be considered the square root of 24 and it is not a perfect square

Answer 25

Method 2: starting from \[3 \sqrt{8} \cdot 4 \sqrt{3}\]

Answer 26

... factor that 8 in the same way, again identifying the one perfect square factor and the one non-perfect-square factor. Could you do that, please?

Answer 27

@12wishs ??

Answer 28

Perfect square=4 and imperfect square=3

Answer 29

Great. Now, what's the square root of 4? of 3?

Answer 30

the square root of 4 is 2 and the square root of 3 is 3

Answer 31

Except that you must write \[\sqrt{3}\]

Answer 32

Also note that Sqrt(24)=Sqrt(4)*Sqrt(6).

Answer 33

\[\sqrt{24}=\sqrt{4}\sqrt{6}\]

Answer 34

or 2\[\sqrt{6}\]

Answer 35

Could you use this result to simplify\[3 \sqrt{8} \cdot 4 \sqrt{3}\]

Answer 36

Note that \[3 \sqrt{8} \cdot 4 \sqrt{3}\]

Answer 37

... can be rearranged as\[24\sqrt{24}=24*2*\sqrt{6}\]

Answer 38

How'd you simplify that?

Answer 39

simplify 6 to its square root of 2 then multiply 24 by 2 which would be 48 then again by 2 which would be 96

Answer 40

Hint: 24*2=?

Answer 41

@mathmale there is something wrong please check it

Answer 42

12sqrt24 not 24sqrt24 i think

Answer 43

I agree that I've stretched this discussion 'way out (without meaning to), and that the problem could be done much faster.

However, we end up with 48 times what? 12wishs?

Answer 44

so wait would the answer to the question be 24 sqrt6?

Answer 45

Yes, indeed. So sorry this has taken so long.

Answer 46

its ok I'm just glad I got an answer

Answer 47

thanks for your help

Answer 48

GREAT @mathmale !!! hope so much that @12wishs have understood it all right now sure

Answer 49

You're welcome. My face is red with embarrassment, however.

Answer 50

;(