complex numbers =D?

Question

is (squareroot) -9 = 9[(squareroot) -1]
???

if its not clear let '~' be square root
~-9 = 9[~-1]

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thanks guys great help

Answer 1

The sqrt (-9) = sqrt (9 * -1) = sqrt (9) * sqrt (-1) =3i.
sqrt(-1) is defined to be i. Also sqrt(9) =3.

You should know that in general:
sqrt(a*b) = sqrt (a) * sqrt(b).
Also sqrt (a/b) = sqrt(a)/sqrt(b)

The answer (3i) is a pure imaginary number becaus it has no real part. A complex number is one that has a real and an imaginary part such as 5+ 6i.

Hope this helped you.

Answer 2

Root(9) is positive 3 by definition.
The solutions of x^2=9 are -3 and +3

Root(-9) = 3i by definition (I am not 100% sure of this one)
The solutions of x^2=-9 are -3i and +3i

Th

Answer 3

Remember, you can break up radical numbers into it's multiples. Therefore:

sqrt(-9) = sqrt(-1)*sqrt(9)

Knowing that i = sqrt(-1), you get:

sqrt(-9) = 3i

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Hope this helps

Answer 4

root (-9) = root(9) * root(-1)

or 3i .....
don't remove root from 9 when you seperate

Answer 5

just so u kno x^.5 = sqrt x

(-9)^.5 = 3i
9*(1)^.5 = 9*i = 9i