Evaluate (6+ square root of 3)^2

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

Step 1

Rewrite ${(6 + \sqrt{3})}^{2}$ as $(6 + \sqrt{3}) (6 + \sqrt{3})$ .

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

Step 2

Expand $(6 + \sqrt{3}) (6 + \sqrt{3})$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Step 3

Simplify and combine like terms.

Simplify each term.

Multiply $6$ by $6$ .

$36 + 6 \sqrt{3} + \sqrt{3} \cdot 6 + \sqrt{3} \sqrt{3}$

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

$36 + 6 \sqrt{3} + 6 \cdot \sqrt{3} + \sqrt{3} \sqrt{3}$

Combine using the product rule for radicals.

$36 + 6 \sqrt{3} + 6 \sqrt{3} + \sqrt{3 \cdot 3}$

Multiply $3$ by $3$ .

$36 + 6 \sqrt{3} + 6 \sqrt{3} + \sqrt{9}$

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

$36 + 6 \sqrt{3} + 6 \sqrt{3} + \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$36 + 6 \sqrt{3} + 6 \sqrt{3} + 3$

Add $36$ and $3$ .

$39 + 6 \sqrt{3} + 6 \sqrt{3}$

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

$39 + 12 \sqrt{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$39 + 12 \sqrt{3}$

Decimal Form:

$59.78460969 \dots$