Simplify the Radical Expression square root of 12+ square root of 108- square root of 32

$\sqrt{12} + \sqrt{108} - \sqrt{32}$

Step 1

Simplify each term.

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$\sqrt{4 (3)} + \sqrt{108} - \sqrt{32}$

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

$\sqrt{2^{2} \cdot 3} + \sqrt{108} - \sqrt{32}$

Pull terms out from under the radical.

$| 2 | \sqrt{3} + \sqrt{108} - \sqrt{32}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$2 \sqrt{3} + \sqrt{108} - \sqrt{32}$

Rewrite $108$ as $6^{2} \cdot 3$ .

Factor $36$ out of $108$ .

$2 \sqrt{3} + \sqrt{36 (3)} - \sqrt{32}$

Rewrite $36$ as $6^{2}$ .

$2 \sqrt{3} + \sqrt{6^{2} \cdot 3} - \sqrt{32}$

Pull terms out from under the radical.

$2 \sqrt{3} + | 6 | \sqrt{3} - \sqrt{32}$

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$2 \sqrt{3} + 6 \sqrt{3} - \sqrt{32}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$2 \sqrt{3} + 6 \sqrt{3} - \sqrt{16 (2)}$

Rewrite $16$ as $4^{2}$ .

$2 \sqrt{3} + 6 \sqrt{3} - \sqrt{4^{2} \cdot 2}$

Pull terms out from under the radical.

$2 \sqrt{3} + 6 \sqrt{3} - (| 4 | \sqrt{2})$

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$2 \sqrt{3} + 6 \sqrt{3} - (4 \sqrt{2})$

Multiply $4$ by $- 1$ .

$2 \sqrt{3} + 6 \sqrt{3} - 4 \sqrt{2}$

Step 2

Add $2 \sqrt{3}$ and $6 \sqrt{3}$ .

$8 \sqrt{3} - 4 \sqrt{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$8 \sqrt{3} - 4 \sqrt{2}$

Decimal Form:

$8.19955221 \dots$