Question

Help me finish please
(sqroot-8) * (sqroot -16)
(sqroot-8)=2i(radical 2), (sqroot -16)=4i
2i(radical 2)*4i= ??? how would i multiply those<<

Answer 1

Square roots are the most common type of radical used. A square root “unsquares”
a number. For example, because 5^2=25 we say the square root of 25 is 5.
The square root of 25 is written as √25.

Answer 2

i know 2i*4i would be -8 but where do i put that into this problem

Answer 3

would it be 2(radical) -8 ?

Answer 4

i is an imaginary number \[i=\sqrt{-1}, for example \sqrt{-16} means \sqrt{16*-1}=\sqrt{16} *\sqrt{-1}=4i , this by product rule \]

Answer 5

I know that..

Answer 6

\[2i(radical 2)*4i= ??? , nothing but 2i \sqrt{2}*4i=2\sqrt{2}*4*i^2, VALUE OF i^2=-1 simplify get your answer\]

Answer 7

so was my answer correct, where would -8 go inside or outside the radical?

Answer 8

outside

Answer 9

\[ProductRule of SquareRoots:\sqrt{a.b}=\sqrt{a}*\sqrt{b}\]

Answer 10

so -8 radical 2

Answer 11

exactly correct

Answer 12

Thankyou