Solve (2/3sqrt{12/3}-sqrt{11/4})^2

$$( \frac { 2 } { 3 } \sqrt { 1 \frac { 2 } { 3 } } - \sqrt { 1 \frac { 1 } { 4 } } ) ^ { 2 }$$

Evaluate

$$-\frac{10\sqrt{3}}{9}+\frac{215}{108}\approx 0.066239843$$

Short Solution Steps

Multiply 1 and 3 to get 3.

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

Add 3 and 2 to get 5.

$$\left(\frac{2}{3}\sqrt{\frac{5}{3}}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

Rewrite the square root of the division \sqrt{\frac{5}{3}} as the division of square roots \frac{\sqrt{5}}{\sqrt{3}}.

$$\left(\frac{2}{3}\times \frac{\sqrt{5}}{\sqrt{3}}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

Rationalize the denominator of \frac{\sqrt{5}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\left(\frac{2}{3}\times \frac{\sqrt{5}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\left(\frac{2}{3}\times \frac{\sqrt{5}\sqrt{3}}{3}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$\left(\frac{2}{3}\times \frac{\sqrt{15}}{3}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

Multiply \frac{2}{3} times \frac{\sqrt{15}}{3} by multiplying numerator times numerator and denominator times denominator.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

Multiply 1 and 4 to get 4.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\sqrt{\frac{4+1}{4}}\right)^{2}$$

Add 4 and 1 to get 5.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\sqrt{\frac{5}{4}}\right)^{2}$$

Rewrite the square root of the division \sqrt{\frac{5}{4}} as the division of square roots \frac{\sqrt{5}}{\sqrt{4}}.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\frac{\sqrt{5}}{\sqrt{4}}\right)^{2}$$

Calculate the square root of 4 and get 2.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\frac{\sqrt{5}}{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\times 3 and 2 is 18. Multiply \frac{2\sqrt{15}}{3\times 3} times \frac{2}{2}. Multiply \frac{\sqrt{5}}{2} times \frac{9}{9}.

$$\left(\frac{2\times 2\sqrt{15}}{18}-\frac{9\sqrt{5}}{18}\right)^{2}$$

Since \frac{2\times 2\sqrt{15}}{18} and \frac{9\sqrt{5}}{18} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{2\times 2\sqrt{15}-9\sqrt{5}}{18}\right)^{2}$$

Do the multiplications in 2\times 2\sqrt{15}-9\sqrt{5}.

$$\left(\frac{4\sqrt{15}-9\sqrt{5}}{18}\right)^{2}$$

To raise \frac{4\sqrt{15}-9\sqrt{5}}{18} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(4\sqrt{15}-9\sqrt{5}\right)^{2}}{18^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4\sqrt{15}-9\sqrt{5}\right)^{2}.

$$\frac{16\left(\sqrt{15}\right)^{2}-72\sqrt{15}\sqrt{5}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

The square of \sqrt{15} is 15.

$$\frac{16\times 15-72\sqrt{15}\sqrt{5}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

Multiply 16 and 15 to get 240.

$$\frac{240-72\sqrt{15}\sqrt{5}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$\frac{240-72\sqrt{5}\sqrt{3}\sqrt{5}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

Multiply \sqrt{5} and \sqrt{5} to get 5.

$$\frac{240-72\times 5\sqrt{3}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

Multiply -72 and 5 to get -360.

$$\frac{240-360\sqrt{3}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{240-360\sqrt{3}+81\times 5}{18^{2}}$$

Multiply 81 and 5 to get 405.

$$\frac{240-360\sqrt{3}+405}{18^{2}}$$

Add 240 and 405 to get 645.

$$\frac{645-360\sqrt{3}}{18^{2}}$$

Calculate 18 to the power of 2 and get 324.

$$\frac{645-360\sqrt{3}}{324}$$

Expand

$$-\frac{10 \sqrt{3}}{9} + \frac{215}{108} = 0.066239843$$

Short Solution Steps

Multiply 1 and 3 to get 3.

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

Add 3 and 2 to get 5.

$$\left(\frac{2}{3}\sqrt{\frac{5}{3}}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

Rewrite the square root of the division \sqrt{\frac{5}{3}} as the division of square roots \frac{\sqrt{5}}{\sqrt{3}}.

$$\left(\frac{2}{3}\times \frac{\sqrt{5}}{\sqrt{3}}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

Rationalize the denominator of \frac{\sqrt{5}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\left(\frac{2}{3}\times \frac{\sqrt{5}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\left(\frac{2}{3}\times \frac{\sqrt{5}\sqrt{3}}{3}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$\left(\frac{2}{3}\times \frac{\sqrt{15}}{3}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

Multiply \frac{2}{3} times \frac{\sqrt{15}}{3} by multiplying numerator times numerator and denominator times denominator.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\sqrt{\frac{1\times 4+1}{4}}\right)^{2}$$

Multiply 1 and 4 to get 4.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\sqrt{\frac{4+1}{4}}\right)^{2}$$

Add 4 and 1 to get 5.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\sqrt{\frac{5}{4}}\right)^{2}$$

Rewrite the square root of the division \sqrt{\frac{5}{4}} as the division of square roots \frac{\sqrt{5}}{\sqrt{4}}.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\frac{\sqrt{5}}{\sqrt{4}}\right)^{2}$$

Calculate the square root of 4 and get 2.

$$\left(\frac{2\sqrt{15}}{3\times 3}-\frac{\sqrt{5}}{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\times 3 and 2 is 18. Multiply \frac{2\sqrt{15}}{3\times 3} times \frac{2}{2}. Multiply \frac{\sqrt{5}}{2} times \frac{9}{9}.

$$\left(\frac{2\times 2\sqrt{15}}{18}-\frac{9\sqrt{5}}{18}\right)^{2}$$

Since \frac{2\times 2\sqrt{15}}{18} and \frac{9\sqrt{5}}{18} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{2\times 2\sqrt{15}-9\sqrt{5}}{18}\right)^{2}$$

Do the multiplications in 2\times 2\sqrt{15}-9\sqrt{5}.

$$\left(\frac{4\sqrt{15}-9\sqrt{5}}{18}\right)^{2}$$

To raise \frac{4\sqrt{15}-9\sqrt{5}}{18} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(4\sqrt{15}-9\sqrt{5}\right)^{2}}{18^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4\sqrt{15}-9\sqrt{5}\right)^{2}.

$$\frac{16\left(\sqrt{15}\right)^{2}-72\sqrt{15}\sqrt{5}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

The square of \sqrt{15} is 15.

$$\frac{16\times 15-72\sqrt{15}\sqrt{5}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

Multiply 16 and 15 to get 240.

$$\frac{240-72\sqrt{15}\sqrt{5}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$\frac{240-72\sqrt{5}\sqrt{3}\sqrt{5}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

Multiply \sqrt{5} and \sqrt{5} to get 5.

$$\frac{240-72\times 5\sqrt{3}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

Multiply -72 and 5 to get -360.

$$\frac{240-360\sqrt{3}+81\left(\sqrt{5}\right)^{2}}{18^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{240-360\sqrt{3}+81\times 5}{18^{2}}$$

Multiply 81 and 5 to get 405.

$$\frac{240-360\sqrt{3}+405}{18^{2}}$$

Add 240 and 405 to get 645.

$$\frac{645-360\sqrt{3}}{18^{2}}$$

Calculate 18 to the power of 2 and get 324.

$$\frac{645-360\sqrt{3}}{324}$$