Find the Product square root of 2( square root of 6+ square root of 7)

$\sqrt{2} (\sqrt{6} + \sqrt{7})$

Simplify terms.

Apply the distributive property.

$\sqrt{2} \sqrt{6} + \sqrt{2} \sqrt{7}$

Combine using the product rule for radicals.

$\sqrt{2 \cdot 6} + \sqrt{2} \sqrt{7}$

Combine using the product rule for radicals.

$\sqrt{2 \cdot 6} + \sqrt{2 \cdot 7}$

$\sqrt{2 \cdot 6} + \sqrt{2 \cdot 7}$

Simplify each term.

Multiply $2$ by $6$ .

$\sqrt{12} + \sqrt{2 \cdot 7}$

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

Factor $4$ out of $12$ .

$\sqrt{4 (3)} + \sqrt{2 \cdot 7}$

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

$\sqrt{2^{2} \cdot 3} + \sqrt{2 \cdot 7}$

$\sqrt{2^{2} \cdot 3} + \sqrt{2 \cdot 7}$

Pull terms out from under the radical.

$2 \sqrt{3} + \sqrt{2 \cdot 7}$

Multiply $2$ by $7$ .

$2 \sqrt{3} + \sqrt{14}$

$2 \sqrt{3} + \sqrt{14}$

The result can be shown in multiple forms.

Exact Form:

$2 \sqrt{3} + \sqrt{14}$

Decimal Form:

$7.20575900 \dots$

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