Solve (sqrt{3}+2/sqrt{3}-frac{sqrt{3}}{2})^2

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

Evaluate

$$\frac{49}{12}\approx 4.083333333$$

Solution Steps

Rationalize the denominator of \frac{2}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Combine \sqrt{3} and \frac{2\sqrt{3}}{3} to get \frac{5}{3}\sqrt{3}.

$$\left(\frac{5}{3}\sqrt{3}-\frac{\sqrt{3}}{2}\right)^{2}$$

Combine \frac{5}{3}\sqrt{3} and -\frac{\sqrt{3}}{2} to get \frac{7}{6}\sqrt{3}.

$$\left(\frac{7}{6}\sqrt{3}\right)^{2}$$

Expand \left(\frac{7}{6}\sqrt{3}\right)^{2}.

$$\left(\frac{7}{6}\right)^{2}\left(\sqrt{3}\right)^{2}$$

Calculate \frac{7}{6} to the power of 2 and get \frac{49}{36}.

$$\frac{49}{36}\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{49}{36}\times 3$$

Multiply \frac{49}{36} and 3 to get \frac{49}{12}.

$$\frac{49}{12}$$

Factor

$$\frac{7 ^ {2}}{2 ^ {2} \cdot 3} = 4\frac{1}{12} = 4.083333333333333$$