Negative Roots

Relevant wiki: Simplifying Expressions with Radicals - Basic

Using complex numbers , we rewrite the expression in terms of $i$ like so: $i\sqrt { 4 } \times i\sqrt { 9 }$ . This leads to ${ i }^{ 2 }\times \sqrt { 36 }$ or $-6$ .

Despite your first instinct, the solution is not $\sqrt { 36 }$ . The answer is not $\sqrt { 36 }$ because you can only multiply the numbers under the radical if they are both positive, as explained here: do square roots always multiply?

$\sqrt { -4 } \sqrt { -9 } \\ =\quad i\sqrt { 4 } (i\sqrt { 9 } )\\ =\quad { i }^{ 2 }\sqrt { 36 } \\ =\quad -1(6)\\ =\quad \boxed { -6 }$

Let me post one thing about this:

1. The definition of $\sqrt{x}$ for $x \in R^+$ is the value of a POSITIVE real number y such that $y^2=x$ .

2. $i = \sqrt{-1}, \sqrt{0}=0$

3. The definition of $\sqrt{-x}$ for $x \in R^+$ is $i$ times the value of a POSITIVE real number y such that $y^2=x$ .

This should be a very basic definition and thus $\sqrt{-4}\sqrt{-9} = (2i)(3i) = \boxed{-6}$

Notes:

1. You can raise a complex number x to the power of non-integer real number only if x is a real number, although I'm not sure that we can raise negative real number to the power of irrational number.

2. This problem is clearly stated and obviously correct. I would also say that it doesn't break the posting rules and guidelines.

3. Most identities for exponentials, on most cases, are unable to be used for complex field. Example: $e^{2\pi i}=1$ , but $e^{\pi i} = -1$ , so $(e^{2\pi i})^{0.5} \neq e^{\pi i}$ .

4. However, $(z^a)^n = z^{an}$ for $z \in C, a, n \in Z, z \neq 0$ still holds. Also for $z \in R^+, a, n \in R$ (although I'm not really sure with this one).

This is incorrect... the answer is there is no answer... there is no square root to a minus number and according to BIDMAS you should square root the numbers before multiplying. Since you can't square root the numbers (this is because squaring a minus number becomes a positive number), you can't reach a solution

I gave the answer as -6 because I knew that only the principal roots are taken here. However, the pair of - 6 and +6 will be a better answer!

$\text{Using simple factoring}\\\sqrt{-4} \sqrt{-9}\\=(\sqrt{4} \times \sqrt{-1}) \times (\sqrt{9} \times \sqrt{-1})\\=2 \times 3 \sqrt{-1}\times \sqrt{-1}\\=6 \times (-1^1)\\=\boxed{-6}$

I think all the arguing is based off of one misunderstanding of math, and a second of how to solve the mystery. Their Misunderstanding: The reason you use + or - in an equation is because of the X term underneath the radical, you don't know if it is a positive or a negative so you need to adjust for that somehow. However, in this problem we know what the number is and therefore do not need the.plus or minus. 2: When solving the problem, the simplest and least complex way is to simplify at the end and not sooner. So you take the square root of both first and end up with 3ii*2i = 6i^2 which implies that it is = -6

It's pretty simple.

sqrt(-4) sqrt(-9)=(i sqrt(4) ) * (i sqrt(9)) Therefore, its (i^2) sqrt(36) which is -1 (+/- 6)

So the answer is +6 or -6

The options are incorrectly labelled. :P

(-4)^1/2 is 2i. (-9)^1/2 is 3i. (2i)(3i)=(2)(3)(-1)=-6

What if both the numbers are equal??? sqrt((-8)^2) = +8 or -8

The answer is -6 and not 6.

Please note that the square root of a number is always positive from the definition of the square root. The only reason we ever have to add a plus/minus sign when dealing with square roots is to preserve a function dealing with variables. note that y=x^2 is not the same as sqrt(y)=x. Hence, in order to preserve all values of the function, we would have to add a plus/minus before sqrt(y). In other words, y=x^2 if and only if x=±sqrt(y).

However, sqrt(4) is not equal to -2. sqrt(4)=2. That's it. If you don't believe me, go to Wolfram Alpha and type "sqrt(4)=-2" into the search box. It will return false. You can also type in sqrt(-4)sqrt(-9)=6 and it will return false.

The squared root of a negative number is the same as the positive number just with i to symbolize it isn't real (squared root of -1 is equal to i) So -4 becomes 2i and -9 becomes 3i (2i)(3i) makes 6i^2 i^2 is -1 as the squared root of anything times itself is the number w/o squared root. So -6 is the final answer after turning 6i^2 into (6)(-1)

although i understand and agree with the solution: -6, because of i squared. Still what the matter of multipling negative radicals as if they where "regular" radicals?

1: sqrt{-4}= sqrt{4} x sqrt{-1} 2: sqrt{-9}=sqrt{9}x sqrt{-1}

sqrt{-4} x sqrt{-9} = sqrt{4} x sqrt{-1} x sqrt{9} x sqrt{-1} = = 2 x 3 x sqrt{-1} x sqrt{-1} = =6 x sqrt{-1}^2 = =6 x (-1) = -6

So, with pemdas, we have to do exponents first (which radicals technically are) so you end up with 2i x 3i which gives 6i^2 which is -6

√-4 = 2i

√-9 = 3i

i^2 = -1

Therefore: (√-4)(√-9) =

2i * 3i = 6i^2 =

6(-1) = -6

The easiest thing would be to start by converting both the numbers into their simplest forms, i.e., their complex forms (OH THE IRONY). We know that:

i = sqrt (-1)

so sqrt (-4) = sqrt (4) * sqrt (-1) => 2i

similarly sqrt (-9) => 3i

now the expression becomes: 2i*3i

2 * 3 = 6, but also:

i * i = sqrt (-1) * sqrt (-1) = sqrt (-1) ^ 2 = -1

Thus: 6 * -1

=> -6 !!

root -4 x root -9 => -1 {root(4x9)} => -1 root 36 => -6