Solve sqrt{-10}*sqrt{-40}

$$\sqrt { - 10 } \cdot \sqrt { - 40 }$$

Evaluate (complex solution)

$$-20$$

Solution Steps

Factor -10=10\left(-1\right). Rewrite the square root of the product \sqrt{10\left(-1\right)} as the product of square roots \sqrt{10}\sqrt{-1}. By definition, the square root of -1 is i.

$$\sqrt{10}i\sqrt{-40}$$

Factor -40=\left(2i\right)^{2}\times 10. Rewrite the square root of the product \sqrt{\left(2i\right)^{2}\times 10} as the product of square roots \sqrt{\left(2i\right)^{2}}\sqrt{10}. Take the square root of \left(2i\right)^{2}.

$$\sqrt{10}i\times \left(2i\right)\sqrt{10}$$

Multiply i and 2i to get -2.

$$\sqrt{10}\left(-2\right)\sqrt{10}$$

Multiply \sqrt{10} and \sqrt{10} to get 10.

$$10\left(-2\right)$$

Multiply 10 and -2 to get -20.

$$-20$$

Real Part (complex solution)

$$-20$$

Solution Steps

Factor -10=10\left(-1\right). Rewrite the square root of the product \sqrt{10\left(-1\right)} as the product of square roots \sqrt{10}\sqrt{-1}. By definition, the square root of -1 is i.

$$Re(\sqrt{10}i\sqrt{-40})$$

Factor -40=\left(2i\right)^{2}\times 10. Rewrite the square root of the product \sqrt{\left(2i\right)^{2}\times 10} as the product of square roots \sqrt{\left(2i\right)^{2}}\sqrt{10}. Take the square root of \left(2i\right)^{2}.

$$Re(\sqrt{10}i\times \left(2i\right)\sqrt{10})$$

Multiply i and 2i to get -2.

$$Re(\sqrt{10}\left(-2\right)\sqrt{10})$$

Multiply \sqrt{10} and \sqrt{10} to get 10.

$$Re(10\left(-2\right))$$

Multiply 10 and -2 to get -20.

$$Re(-20)$$

The real part of -20 is -20.

$$-20$$

Evaluate

$$\text{Indeterminate}$$