Evaluate square root of 12+ square root of 48- square root of 27

$\sqrt{12} + \sqrt{48} - \sqrt{27}$

Step 1

Simplify each term.

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

Factor $4$ out of $12$ .

$\sqrt{4 (3)} + \sqrt{48} - \sqrt{27}$

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

$\sqrt{2^{2} \cdot 3} + \sqrt{48} - \sqrt{27}$

$\sqrt{2^{2} \cdot 3} + \sqrt{48} - \sqrt{27}$

Pull terms out from under the radical.

$2 \sqrt{3} + \sqrt{48} - \sqrt{27}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$2 \sqrt{3} + \sqrt{16 (3)} - \sqrt{27}$

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

$2 \sqrt{3} + \sqrt{4^{2} \cdot 3} - \sqrt{27}$

$2 \sqrt{3} + \sqrt{4^{2} \cdot 3} - \sqrt{27}$

Pull terms out from under the radical.

$2 \sqrt{3} + 4 \sqrt{3} - \sqrt{27}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$2 \sqrt{3} + 4 \sqrt{3} - \sqrt{9 (3)}$

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

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

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

Pull terms out from under the radical.

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

Multiply $3$ by $- 1$ .

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

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

Step 2

Simplify by adding terms.

Add $2 \sqrt{3}$ and $4 \sqrt{3}$ .

$6 \sqrt{3} - 3 \sqrt{3}$

Subtract $3 \sqrt{3}$ from $6 \sqrt{3}$ .

$3 \sqrt{3}$

$3 \sqrt{3}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$3 \sqrt{3}$

Decimal Form:

$5.19615242 \dots$

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