Evaluate square root of 6(3+ square root of 2)

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

Step 1

Simplify terms.

Apply the distributive property.

$\sqrt{6} \cdot 3 + \sqrt{6} \sqrt{2}$

Move $3$ to the left of $\sqrt{6}$ .

$3 \cdot \sqrt{6} + \sqrt{6} \sqrt{2}$

Combine using the product rule for radicals.

$3 \cdot \sqrt{6} + \sqrt{6 \cdot 2}$

$3 \cdot \sqrt{6} + \sqrt{6 \cdot 2}$

Step 2

Simplify each term.

Multiply $6$ by $2$ .

$3 \sqrt{6} + \sqrt{12}$

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

Factor $4$ out of $12$ .

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

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

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

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

Pull terms out from under the radical.

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

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

Step 3

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$10.81257084 \dots$

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