Solve (sqrt{8}+i)(sqrt{8}-i)

$$( \sqrt { 8 } + i ) ( \sqrt { 8 } - i )$$

Evaluate

$$9$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}+i\right)\left(2\sqrt{2}-i\right)$$

Apply the distributive property by multiplying each term of 2\sqrt{2}+i by each term of 2\sqrt{2}-i.

$$4\left(\sqrt{2}\right)^{2}-2i\sqrt{2}+2i\sqrt{2}+1$$

The square of \sqrt{2} is 2.

$$4\times 2-2i\sqrt{2}+2i\sqrt{2}+1$$

Multiply 4 and 2 to get 8.

$$8-2i\sqrt{2}+2i\sqrt{2}+1$$

Combine -2i\sqrt{2} and 2i\sqrt{2} to get 0.

$$8+1$$

Add 8 and 1 to get 9.

$$9$$

Real Part

$$9$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$Re(\left(2\sqrt{2}+i\right)\left(\sqrt{8}-i\right))$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Apply the distributive property by multiplying each term of 2\sqrt{2}+i by each term of 2\sqrt{2}-i.

$$Re(4\left(\sqrt{2}\right)^{2}-2i\sqrt{2}+2i\sqrt{2}+1)$$

The square of \sqrt{2} is 2.

$$Re(4\times 2-2i\sqrt{2}+2i\sqrt{2}+1)$$

Multiply 4 and 2 to get 8.

$$Re(8-2i\sqrt{2}+2i\sqrt{2}+1)$$

Combine -2i\sqrt{2} and 2i\sqrt{2} to get 0.

$$Re(8+1)$$

Add 8 and 1 to get 9.

$$Re(9)$$

The real part of 9 is 9.

$$9$$