Ze basics

Obviously, it's $\sqrt{6}$ . Well, guess what? $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ doesn't hold true when $a,b<0$ . So we bring in complex numbers.

$\sqrt{-2}\sqrt{-3}=\left(\sqrt{-1}\right)\left(\sqrt{2}\right)\left(\sqrt{-1}\right)\left(\sqrt{3}\right)=(i)\left(\sqrt{2}\right)(i)\left(\sqrt{3}\right)=\boxed{-\sqrt{6}}$

Learned this in math class today!

Note: This solution is meant to be as newbie-friendly as possible. In fact, I didn't learn this in math class; I just browsed about the imaginary number sometime ago. I assume that you have understood the 4 basic number operations, integers, exponents, roots, and variables.

Algebraic Notes:

1. $-1 * a = -a$ (The rule stating that anything multiplied by one is equal to itself combined with the rule stating that multiplication of two numbers with different signs will result in a negative product.)

2. $i = \sqrt{-1}$ (Impossible to solve, this is called the imaginary number.)

3. $\sqrt{a} * \sqrt{a} = (\sqrt{a})^2 = a$ (Exponentiation first. Since the power of the root and the exponent is then just equal but opposite, cancel out both.)

4. $i^2 = (\sqrt{-1})^2 = -1$ (Combination of rule 2 and 3 stated above. If you try to square the imaginary number, you will square the square root of $-1$ , which will then cancel out and result in $-1$ .)

5. if $a, b > 0$ , or $a = -1$ and $b > 0$ , or $a > 0$ and $b = -1$ , then $\sqrt{ab} = \sqrt{a} * \sqrt{b}$ (Given two positive numbers or either one being the imaginary number, the root of a product is equal to the root of the multiplicand multiplied by the root of the multiplier.)

6. $ab = ba$ (The commutative property of multiplication. Applies to addition, too.)

Solution:

Use the imaginary number, some square root algebra and some basic algebra shown above. It is solved below.

$\sqrt{-2} * \sqrt{-3}$

$= \sqrt{2 * -1} * \sqrt{3 * -1}$

$= \sqrt{2} * \sqrt{-1} * \sqrt{3} * \sqrt{-1}$

$= \sqrt{2} * i * \sqrt{3} * i$

$= i * \sqrt{2} * \sqrt{3} * i$

$= i * \sqrt{2} * i * \sqrt{3}$

$= i * i * \sqrt{2} * \sqrt{3}$

$= i^2 * \sqrt{2} * \sqrt{3}$

$= -1 * \sqrt{2} * \sqrt{3}$

$= -1 * \sqrt{2 * 3}$

$= -1 * \sqrt{6}$

$= -\sqrt{6}$

Correct Answer: $-\sqrt{6}$

$\sqrt { -3 } \cdot \sqrt { -2 } \\ =\sqrt { 3 } i\cdot \sqrt { 2 } i\\ =\sqrt { 6 } { i }^{ 2 }\\ =-\sqrt { 6 }$

The square root function is double-valued over the complex plane, and there is no good motivation for picking one sheet and ignoring the other. I suggest removing this problem, since it has no unambiguous answer.

Recall that $i^2=-1$ so $-2=2 \times -1=2 \times i^2$ and similarly for $-3=3 \times i^2$ so $\sqrt{-2}=i\sqrt{2}$ and $\sqrt{-3} =i\sqrt{3}$ hence $\sqrt{-2} \times \sqrt{-3} = i\sqrt{2} \times i\sqrt{3}$ $= i^2 \sqrt{3 \times 2} =i^2 \sqrt{6} =-\sqrt{6}$

A common problem seems to be understanding when $\sqrt a\sqrt b=\sqrt{ab}$ holds true.

Firstly: $\left(\pm\sqrt a\sqrt b\right)^2=\left(\pm\sqrt{ab}\right)^2=ab$ , for any $a,\,b\in\mathbb C$ .

With any number $z\in\mathbb C$ , there are two “square roots”, namely $\pm\sqrt z$ . When $z$ is real and non-negative, we choose the real non-negative root by convention: so $\sqrt4=2$ , not $-2$ (even though $(-2)^2=4$ just as much as $2^2=4$ ). Therefore, when $a$ and $b$ are real and non-negative, $\sqrt a$ and $\sqrt b$ are real and non-negative and so, finally, is their product. This means that $\sqrt a\sqrt b=\sqrt{ab}$ for any real non-negative $a$ and $b$ .

The trouble comes with complex numbers because of which of the two possible roots (in $\mathbb C$ ) we choose by convention. When we fix $\sqrt{-1}=i$ , we then decide that $\sqrt{-x}$ means $\sqrt{x}\,i$ , not $-\sqrt x\,i$ (even though both square to give $-x$ ), for any real negative number $-x$ . Unfortunately, this means that when we multiply two square roots of negatives, $\sqrt{-x}$ and $\sqrt{-y}$ , the following happens:

$\sqrt{-x}\sqrt{-y}\\ =\sqrt{x}\,i\times\sqrt{y}\,i\\ =i^2\times\sqrt{x}\sqrt{y}\\ =-\sqrt{xy},$

so that $\sqrt a\sqrt b=\sqrt{ab}$ is not (quite) true. Note: the third–fourth line is allowed, because $x$ and $y$ are real and non-negative, as above. Also: see the first point to see that this is only a problem with notation, not with maths...

In conclusion: only trust square roots (“ $\sqrt x$ ” notation) for non-negative real numbers. Otherwise, all hail $i$ . And use indices.

2^(1/2)ix3^(1/2)i 6^(1/2)x(-1) -6^(1/2)

$\sqrt{-2} \times \sqrt{-3}=i\sqrt{2} \times i\sqrt{3}=i \times i \times \sqrt{2} \times \sqrt{3}=i^2 \times \sqrt{2 \times 3}=-1 \times \sqrt{6}=\boxed{\large{-\sqrt{6}}}$

By definition we have that $i^{2} = -1$ , so $\sqrt{-2} \times \sqrt{-3} = \sqrt{2i^{2}} \times \sqrt{3i^{2}} = i\sqrt{2} \times i\sqrt{3} = i^{2}\sqrt{6} = -1\sqrt{6} = -\sqrt{6}$ .

$\large \sqrt[4]{-2} \times \sqrt[4]{-3} \\\large= \sqrt[4]{-1} \times \sqrt[4]{2}\times\large \sqrt[4]{-1} \times \sqrt[4]{3} \\ \large= \sqrt[4]{-1} \times \sqrt[4]{-1}\times \sqrt[4]{2} \times \sqrt[4]{3} \\ \large = \sqrt[2]{-1} \times \sqrt[4]{6} \\\large = i\sqrt[4]{6}$

root (-1) = i therefore it will be 2i*3i = 6i^2 = -6

Do it stepwise. First multiple the numbers within the roots. √2 * √3 = √6

Now multiply the signs, √- * √- = ( √- )^2 = -

Combining, Answer is -√6

((-1) (2))^.5 ((-1) (3))^.5=i (2^.5)*i(3^.5)=i^2(6^.5)=-6^.5

Actually there is no ambiguity in the.answer if you represent as.you.should a complex numbe z as |z|exp(i phi), with phi the phase of such a number and |z| its modulos.

The radical precedes multiplication in the order of operations. The principle square root of $-2$ is $i \sqrt{2}$ and the principle square root of $-3$ is $i \sqrt{3}$ therefore their product is $i\times i \times \sqrt{6} = -\sqrt{6}$

As a, b < 0. So √a √b ≠ √ab now √-2 = i√2 and √-3 = i√3 this imply (√-2 ) (√-3) = $i^{2}$ (√2)(√3) = -√6 where i is an imaginary unit defined as i = √-1

$\Large \sqrt{-2} \times \sqrt{-3} = \sqrt {2}i \times \sqrt {3}i$ $\Large = \sqrt{6}i^{2}$ Remember that $\Large i^{2}=-1$ $\Large \therefore \Large \sqrt{6}i^{2}=\sqrt{6}\times (-1)$ $\Large=-\sqrt{6}$

√2i × √3i = √6i^2 which gives -√6 as i^2 = -1

$\sqrt{-2}\times \sqrt{-3}=i\sqrt{2}\times i\sqrt{3}$ Then $i^{2}=-1$ So $(-1)\times 2^{\frac{1}{2}}\times 3^{\frac{1}{2}}=(-1) \times 6^{\frac{1}{2}}=-\sqrt{6}$

√a×√b=√ab Only when at least one of them is greater than zero. Here both -2 and -3 are less than zero, hence there is only one answer. √(-2)×√(-3)=√2 i×√3 i=√6 i^2=-√6