Solve sqrt{2}div1/sqrt{3+1}*(sqrt{3}+1)

$$\sqrt { 2 } \div \frac { 1 } { \sqrt { 3 } + 1 } \times ( \sqrt { 3 } + 1 )$$

Evaluate

$$2\sqrt{2}\left(\sqrt{3}+2\right)\approx 10.555833735$$

Solution Steps

Divide \sqrt{2} by \frac{1}{\sqrt{3}+1} by multiplying \sqrt{2} by the reciprocal of \frac{1}{\sqrt{3}+1}.

$$\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}+1\right)$$

Use the distributive property to multiply \sqrt{2} by \sqrt{3}+1.

$$\left(\sqrt{2}\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+1\right)$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}+1\right)$$

Apply the distributive property by multiplying each term of \sqrt{6}+\sqrt{2} by each term of \sqrt{3}+1.

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

Combine \sqrt{6} and \sqrt{6} to get 2\sqrt{6}.

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

Combine 3\sqrt{2} and \sqrt{2} to get 4\sqrt{2}.

$$4\sqrt{2}+2\sqrt{6}$$