Find the Complex Conjugate square root of 8- square root of 9

$\sqrt{8} - \sqrt{9}$

Simplify each term.

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$\sqrt{4 (2)} - \sqrt{9}$

Rewrite $4$ as $2^{2}$ .

$\sqrt{2^{2} \cdot 2} - \sqrt{9}$

Pull terms out from under the radical.

$2 \sqrt{2} - \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

$2 \sqrt{2} - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$2 \sqrt{2} - 3$

Since there are no imaginary terms, the complex conjugate is the same as the simplified expression.

$2 \sqrt{2} - 3$

The result can be shown in multiple forms.

Exact Form:

$2 \sqrt{2} - 3$

Decimal Form:

$- 0.17157287 \dots$