Question

√(-108)+√(-27)

√(-6)∙√(-18)

Answer 1

\[\sqrt{-108} + \sqrt{-27}\]

\[\sqrt{-6} * \sqrt{-18}\]

Answer 2

is this 2 different problems ?

Answer 3

@phi I think so.

Answer 4

yes

Answer 5

@phi @vzfreakz

Answer 6

Do you have a scientific calculator?

That'd be the easy way. If not, then that's fine.

Answer 7

yes i do

Answer 8

the first thing to do is factor the numbers inside the square root.
-108 is 4* -27

\[ \sqrt{ 4 \cdot 27 \cdot -1 } = \sqrt{4} \sqrt{27} i= 2 \sqrt{27} i\]

Answer 9

so the first problem is
\[ 2 \sqrt{27} i+ \sqrt{27} i \]
which can be simplified. (btw, i is short for sqr(-1) )

Answer 10

how would you simplify

Answer 11

you have 2 of sqr(27) * i plus one more sqr(27)*i
how many sqr(27)*i do you have ?

Answer 12

you have 3 of them

Answer 13

\[3i \sqrt{15}\]

Answer 14

?

Answer 15

except the 15 should be 27

Answer 16

oh...that dosnt match any of the answers

Answer 17

are you sure you have the correct question?

or we may be dealing with a typo in the question or answer choices.
what are the other choices?

Answer 18

question \[\sqrt{-108}+\sqrt{-27}\]
\[-3\sqrt{15}\]
\[3i \sqrt{15}\]
\[9i \sqrt{3}\]
\[-9\sqrt{3}\]

Answer 19

ok. so far we have
\[ 3 i \sqrt{27} \]

we can do more. first, can you factor 27 ?

Answer 20

3,3,3?

Answer 21

yes, 3*3*3

we can simplify a PAIR inside a square root.
in other words
\[ \sqrt{3\cdot 3 \cdot 3 } = \sqrt{3\cdot3} \cdot \sqrt{3} \]

Answer 22

remember, for any number "a" we can simplify the sqr( a*a)
\[ \sqrt{a\cdot a} =a \]

Answer 23

ok

Answer 24

\[3\sqrt{3}\]

Answer 25

@phi

Answer 26

yes, so
\[ 3i \sqrt{27} = 3 i \cdot 3 \sqrt{3} \]
you can simplify the 3*3 part