Solve (8+sqrt{-6})^2

$$( 8 + \sqrt { - 6 } ) ^ { 2 }$$

Evaluate (complex solution)

$$58+16\sqrt{6}i\approx 58+39.191835885i$$

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.

$$\left(8+\sqrt{6}i\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8+\sqrt{6}i\right)^{2}.

$$64+16i\sqrt{6}-\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$64+16i\sqrt{6}-6$$

Subtract 6 from 64 to get 58.

$$58+16i\sqrt{6}$$

Expand (complex solution)

$$58+16\sqrt{6}i$$

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.

$$\left(8+\sqrt{6}i\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8+\sqrt{6}i\right)^{2}.

$$64+16i\sqrt{6}-\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$64+16i\sqrt{6}-6$$

Subtract 6 from 64 to get 58.

$$58+16i\sqrt{6}$$

Evaluate

$$\text{Indeterminate}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8+\sqrt{-6}\right)^{2}.

$$64+16\sqrt{-6}+\left(\sqrt{-6}\right)^{2}$$

Calculate \sqrt{-6} to the power of 2 and get -6.

$$64+16\sqrt{-6}-6$$

Subtract 6 from 64 to get 58.

$$58+16\sqrt{-6}$$