Pre-Algebra Examples

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

Step 1

Simplify terms.

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Step 1.1

Apply the distributive property.

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

Step 1.2

Combine using the product rule for radicals.

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

Step 1.3

Combine using the product rule for radicals.

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

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

Step 2

Simplify each term.

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Step 2.1

Multiply $6$ by $2$ .

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

Step 2.2

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

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Step 2.2.1

Factor $4$ out of $12$ .

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

Step 2.2.2

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

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

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

Step 2.3

Pull terms out from under the radical.

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

Step 2.4

Multiply $6$ by $3$ .

$2 \sqrt{3} + \sqrt{18}$

Step 2.5

Rewrite $18$ as $3^{2} \cdot 2$ .

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Step 2.5.1

Factor $9$ out of $18$ .

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

Step 2.5.2

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

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

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

Step 2.6

Pull terms out from under the radical.

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

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

Step 3

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$7.70674230 \dots$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$