Simplify (3+ square root of 2)^2

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

Step 1

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

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

Step 2

Expand $(3 + \sqrt{2}) (3 + \sqrt{2})$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Step 3

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$9 + 3 \sqrt{2} + \sqrt{2} \cdot 3 + \sqrt{2} \sqrt{2}$

Move $3$ to the left of $\sqrt{2}$ .

$9 + 3 \sqrt{2} + 3 \cdot \sqrt{2} + \sqrt{2} \sqrt{2}$

Combine using the product rule for radicals.

$9 + 3 \sqrt{2} + 3 \sqrt{2} + \sqrt{2 \cdot 2}$

Multiply $2$ by $2$ .

$9 + 3 \sqrt{2} + 3 \sqrt{2} + \sqrt{4}$

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

$9 + 3 \sqrt{2} + 3 \sqrt{2} + \sqrt{2^{2}}$

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

$9 + 3 \sqrt{2} + 3 \sqrt{2} + 2$

Add $9$ and $2$ .

$11 + 3 \sqrt{2} + 3 \sqrt{2}$

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

$11 + 6 \sqrt{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$11 + 6 \sqrt{2}$

Decimal Form:

$19.48528137 \dots$