Solve frac{sqrt{-6}sqrt{-2}}{sqrt{3}}

$$\frac{ \sqrt{ -6 } \sqrt{ -2 } }{ \sqrt{ 3 } }$$

Evaluate (complex solution)

$$2$$

Solution Steps

Factor -6=6\left(-1\right). Rewrite the square root of the product \sqrt{6\left(-1\right)} as the product of square roots \sqrt{6}\sqrt{-1}. By definition, the square root of -1 is i.

$$\frac{\sqrt{6}i\sqrt{-2}}{\sqrt{3}}$$

Factor -2=2\left(-1\right). Rewrite the square root of the product \sqrt{2\left(-1\right)} as the product of square roots \sqrt{2}\sqrt{-1}. By definition, the square root of -1 is i.

$$\frac{\sqrt{6}i\sqrt{2}i}{\sqrt{3}}$$

Multiply i and i to get -1.

$$\frac{\sqrt{6}\left(-1\right)\sqrt{2}}{\sqrt{3}}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{\sqrt{2}\sqrt{3}\left(-1\right)\sqrt{2}}{\sqrt{3}}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{2\left(-1\right)\sqrt{3}}{\sqrt{3}}$$

Rationalize the denominator of \frac{2\left(-1\right)\sqrt{3}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{2\left(-1\right)\sqrt{3}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{2\left(-1\right)\sqrt{3}\sqrt{3}}{3}$$

Multiply 2 and -1 to get -2.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{-2\times 3}{3}$$

Multiply -2 and 3 to get -6.

$$\frac{-6}{3}$$

Divide -6 by 3 to get -2.

$$-2$$

Real Part (complex solution)

$$2$$

Solution Steps

Factor -6=6\left(-1\right). Rewrite the square root of the product \sqrt{6\left(-1\right)} as the product of square roots \sqrt{6}\sqrt{-1}. By definition, the square root of -1 is i.

$$Re(\frac{\sqrt{6}i\sqrt{-2}}{\sqrt{3}})$$

Factor -2=2\left(-1\right). Rewrite the square root of the product \sqrt{2\left(-1\right)} as the product of square roots \sqrt{2}\sqrt{-1}. By definition, the square root of -1 is i.

$$Re(\frac{\sqrt{6}i\sqrt{2}i}{\sqrt{3}})$$

Multiply i and i to get -1.

$$Re(\frac{\sqrt{6}\left(-1\right)\sqrt{2}}{\sqrt{3}})$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$Re(\frac{\sqrt{2}\sqrt{3}\left(-1\right)\sqrt{2}}{\sqrt{3}})$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$Re(\frac{2\left(-1\right)\sqrt{3}}{\sqrt{3}})$$

Rationalize the denominator of \frac{2\left(-1\right)\sqrt{3}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$Re(\frac{2\left(-1\right)\sqrt{3}\sqrt{3}}{\left(\sqrt{3}\right)^{2}})$$

The square of \sqrt{3} is 3.

$$Re(\frac{2\left(-1\right)\sqrt{3}\sqrt{3}}{3})$$

Multiply 2 and -1 to get -2.

$$Re(\frac{-2\sqrt{3}\sqrt{3}}{3})$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$Re(\frac{-2\times 3}{3})$$

Multiply -2 and 3 to get -6.

$$Re(\frac{-6}{3})$$

Divide -6 by 3 to get -2.

$$Re(-2)$$

The real part of -2 is -2.

$$-2$$

Evaluate

$$\text{Indeterminate}$$