Evaluate square root of 6(2 square root of 10+3 square root of 2)

$\sqrt{6} (2 \sqrt{10} + 3 \sqrt{2})$

Step 1

Apply the distributive property.

$\sqrt{6} (2 \sqrt{10}) + \sqrt{6} (3 \sqrt{2})$

Step 2

Multiply $\sqrt{6} (2 \sqrt{10})$ .

Combine using the product rule for radicals.

$2 \sqrt{6 \cdot 10} + \sqrt{6} (3 \sqrt{2})$

Multiply $6$ by $10$ .

$2 \sqrt{60} + \sqrt{6} (3 \sqrt{2})$

$2 \sqrt{60} + \sqrt{6} (3 \sqrt{2})$

Step 3

Multiply $\sqrt{6} (3 \sqrt{2})$ .

Combine using the product rule for radicals.

$2 \sqrt{60} + 3 \sqrt{6 \cdot 2}$

Multiply $6$ by $2$ .

$2 \sqrt{60} + 3 \sqrt{12}$

$2 \sqrt{60} + 3 \sqrt{12}$

Step 4

Simplify each term.

Rewrite $60$ as $2^{2} \cdot 15$ .

Factor $4$ out of $60$ .

$2 \sqrt{4 (15)} + 3 \sqrt{12}$

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

$2 \sqrt{2^{2} \cdot 15} + 3 \sqrt{12}$

$2 \sqrt{2^{2} \cdot 15} + 3 \sqrt{12}$

Pull terms out from under the radical.

$2 (2 \sqrt{15}) + 3 \sqrt{12}$

Multiply $2$ by $2$ .

$4 \sqrt{15} + 3 \sqrt{12}$

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

Factor $4$ out of $12$ .

$4 \sqrt{15} + 3 \sqrt{4 (3)}$

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

$4 \sqrt{15} + 3 \sqrt{2^{2} \cdot 3}$

$4 \sqrt{15} + 3 \sqrt{2^{2} \cdot 3}$

Pull terms out from under the radical.

$4 \sqrt{15} + 3 (2 \sqrt{3})$

Multiply $2$ by $3$ .

$4 \sqrt{15} + 6 \sqrt{3}$

$4 \sqrt{15} + 6 \sqrt{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$4 \sqrt{15} + 6 \sqrt{3}$

Decimal Form:

$25.88423823 \dots$

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