Solve 2/sqrt{2}(2sqrt{12}+4sqrt{1/8}-3)

$$\frac { 2 } { \sqrt { 2 } } ( 2 \sqrt { 12 } + 4 \sqrt { \frac { 1 } { 8 } } - 3 )$$

Evaluate

$$4\sqrt{6}+2-3\sqrt{2}\approx 7.555318284$$

Solution Steps

Rationalize the denominator of \frac{2}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{2\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\left(2\sqrt{12}+4\sqrt{\frac{1}{8}}-3\right)$$

The square of \sqrt{2} is 2.

$$\frac{2\sqrt{2}}{2}\left(2\sqrt{12}+4\sqrt{\frac{1}{8}}-3\right)$$

Cancel out 2 and 2.

$$\sqrt{2}\left(2\sqrt{12}+4\sqrt{\frac{1}{8}}-3\right)$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\sqrt{2}\left(2\times 2\sqrt{3}+4\sqrt{\frac{1}{8}}-3\right)$$

Multiply 2 and 2 to get 4.

$$\sqrt{2}\left(4\sqrt{3}+4\sqrt{\frac{1}{8}}-3\right)$$

Rewrite the square root of the division \sqrt{\frac{1}{8}} as the division of square roots \frac{\sqrt{1}}{\sqrt{8}}.

$$\sqrt{2}\left(4\sqrt{3}+4\times \frac{\sqrt{1}}{\sqrt{8}}-3\right)$$

Calculate the square root of 1 and get 1.

$$\sqrt{2}\left(4\sqrt{3}+4\times \frac{1}{\sqrt{8}}-3\right)$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\sqrt{2}\left(4\sqrt{3}+4\times \frac{1}{2\sqrt{2}}-3\right)$$

Rationalize the denominator of \frac{1}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\sqrt{2}\left(4\sqrt{3}+4\times \frac{\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}-3\right)$$

The square of \sqrt{2} is 2.

$$\sqrt{2}\left(4\sqrt{3}+4\times \frac{\sqrt{2}}{2\times 2}-3\right)$$

Multiply 2 and 2 to get 4.

$$\sqrt{2}\left(4\sqrt{3}+4\times \frac{\sqrt{2}}{4}-3\right)$$

Cancel out 4 and 4.

$$\sqrt{2}\left(4\sqrt{3}+\sqrt{2}-3\right)$$

Use the distributive property to multiply \sqrt{2} by 4\sqrt{3}+\sqrt{2}-3.

$$4\sqrt{2}\sqrt{3}+\left(\sqrt{2}\right)^{2}-3\sqrt{2}$$

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

$$4\sqrt{6}+\left(\sqrt{2}\right)^{2}-3\sqrt{2}$$

The square of \sqrt{2} is 2.

$$4\sqrt{6}+2-3\sqrt{2}$$