Question

If a and b are both positive real no.s, which solution for this: (Sqrt(-a))(sqrt(-b))
Is correct?

1st sol:
Sqrt(a) times sqrt(b) times i squared
= -sqrt(ab)

2nd sol:

Sqrt((-a)(-b))
=sqrt(ab)

Could someone please explain this to me? Thanks a lot. Help is much appreciated.

Answer 1

hint: \[\sqrt{-1}=i\]

Answer 2

I know. But why does seperating the i from-a and -b before multiplying gives a different answer than just multiplying -a and -b before sqrt-ing?

Answer 3

In simplest terms, you can only multiply them on the inside IF AND ONLY IF they are both positive

Answer 4

ok, thanks.

Is there any way to prove that this method is true and the other is not?
Or why in particular this is the way to do it and not the other?

Answer 5

Honestly I don't remember how I used to rationalize it. I've justkinda accepted it by now. mathstudent55's answer should be good, though

Answer 6

Im actually long past the stage of learning complex numbers, but then a few days ago, my little brother stumped me with this q, and I didn't know how to explain it to him. So, I knew I have to ask ppl for help.

Answer 7

Using only real numbers, this problem has no solution.

\(\sqrt{-4} \times \sqrt{-16} \)

The reason is that \(\sqrt{-4}\) and \(\sqrt{-16} \) are not defined for real numbers.

Once you allow imaginary numbers, then you must deal with the square root of -1 being equal to i.

Then the problem

\(\sqrt{-4} \times \sqrt{-16} \)

is solved this way and has a solution:

\(= i\sqrt{4} \times i\sqrt{16} \)

\(= 2i \times 4i\)

\(= 8i^2\)

\(= -8\)

Answer 8

oh... Thanks a million. :)

Answer 9

Thanks everybody, help is greatly appreciated.

Answer 10

You're welcome.

Answer 11

@mathstudent55 I got one for ya. Simplify:
\[i\sqrt{-12} - \sqrt{12}\]

Answer 12

-4sqrt3?

Answer 13

I'll do it one step at a time. You can do this in much fewer steps, but I'm showing every step.

\(i\sqrt{-12} - \sqrt{12}\)

\(= i\sqrt{-1 \times 12} - \sqrt{12}\)

\(= i\sqrt{-1} \times \sqrt{12} - \sqrt{12}\)

\(= i\cdot i \times \sqrt{12} - \sqrt{12}\)

\(= -1 \times \sqrt{12} - \sqrt{12}\)

\(= -\sqrt{12} - \sqrt{12}\)

\(= -2\sqrt{12}\)

\(= -2\sqrt{4 \times 3}\)

\(= -2\sqrt{4} \sqrt{3}\)

\(= -2\times 2 \sqrt{3}\)

\(= -4 \sqrt{3}\)

@tiffanymak1966 is correct