Solve (frac{sqrt{6}}{2}+sqrt{2})^2+2/4

$$( \frac { \sqrt { 6 } } { 2 } + \sqrt { 2 } ) ^ { 2 } + \frac { 2 } { 4 }$$

Evaluate

$$2\left(\sqrt{3}+2\right)\approx 7.464101615$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{2} times \frac{2}{2}.

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

Since \frac{\sqrt{6}}{2} and \frac{2\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

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

To raise \frac{\sqrt{6}+2\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2^{2} and 2 is 4. Multiply \frac{1}{2} times \frac{2}{2}.

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

Since \frac{\left(\sqrt{6}+2\sqrt{2}\right)^{2}}{4} and \frac{2}{4} have the same denominator, add them by adding their numerators.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+2\sqrt{2}\right)^{2}.

$$\frac{\left(\sqrt{6}\right)^{2}+4\sqrt{6}\sqrt{2}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

The square of \sqrt{6} is 6.

$$\frac{6+4\sqrt{6}\sqrt{2}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

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

$$\frac{6+4\sqrt{2}\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{6+4\times 2\sqrt{3}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

Multiply 4 and 2 to get 8.

$$\frac{6+8\sqrt{3}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

The square of \sqrt{2} is 2.

$$\frac{6+8\sqrt{3}+4\times 2}{2^{2}}+\frac{1}{2}$$

Multiply 4 and 2 to get 8.

$$\frac{6+8\sqrt{3}+8}{2^{2}}+\frac{1}{2}$$

Add 6 and 8 to get 14.

$$\frac{14+8\sqrt{3}}{2^{2}}+\frac{1}{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{14+8\sqrt{3}}{4}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{1}{2} times \frac{2}{2}.

$$\frac{14+8\sqrt{3}}{4}+\frac{2}{4}$$

Since \frac{14+8\sqrt{3}}{4} and \frac{2}{4} have the same denominator, add them by adding their numerators.

$$\frac{14+8\sqrt{3}+2}{4}$$

Do the calculations in 14+8\sqrt{3}+2.

$$\frac{16+8\sqrt{3}}{4}$$

Expand

$$2 \sqrt{3} + 4 = 7.464101615$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{2} times \frac{2}{2}.

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

Since \frac{\sqrt{6}}{2} and \frac{2\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

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

To raise \frac{\sqrt{6}+2\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2^{2} and 2 is 4. Multiply \frac{1}{2} times \frac{2}{2}.

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

Since \frac{\left(\sqrt{6}+2\sqrt{2}\right)^{2}}{4} and \frac{2}{4} have the same denominator, add them by adding their numerators.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+2\sqrt{2}\right)^{2}.

$$\frac{\left(\sqrt{6}\right)^{2}+4\sqrt{6}\sqrt{2}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

The square of \sqrt{6} is 6.

$$\frac{6+4\sqrt{6}\sqrt{2}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

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

$$\frac{6+4\sqrt{2}\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{6+4\times 2\sqrt{3}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

Multiply 4 and 2 to get 8.

$$\frac{6+8\sqrt{3}+4\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{1}{2}$$

The square of \sqrt{2} is 2.

$$\frac{6+8\sqrt{3}+4\times 2}{2^{2}}+\frac{1}{2}$$

Multiply 4 and 2 to get 8.

$$\frac{6+8\sqrt{3}+8}{2^{2}}+\frac{1}{2}$$

Add 6 and 8 to get 14.

$$\frac{14+8\sqrt{3}}{2^{2}}+\frac{1}{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{14+8\sqrt{3}}{4}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{1}{2} times \frac{2}{2}.

$$\frac{14+8\sqrt{3}}{4}+\frac{2}{4}$$

Since \frac{14+8\sqrt{3}}{4} and \frac{2}{4} have the same denominator, add them by adding their numerators.

$$\frac{14+8\sqrt{3}+2}{4}$$

Do the calculations in 14+8\sqrt{3}+2.

$$\frac{16+8\sqrt{3}}{4}$$