Solve sqrt{{left(frac{9sqrt{3}}{2}right)}^2-{left(frac{3sqrt{3}}{2}right)}^2}

$$\sqrt{ { \left( \frac{ 9 \sqrt{ 3 } }{ 2 } \right) }^{ 2 } - { \left( \frac{ 3 \sqrt{ 3 } }{ 2 } \right) }^{ 2 } }$$

Evaluate

$$3\sqrt{6}\approx 7.348469228$$

Solution Steps

To raise \frac{9\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}}{2^{2}}-\left(\frac{3\sqrt{3}}{2}\right)^{2}}$$

To raise \frac{3\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}}{2^{2}}-\frac{\left(3\sqrt{3}\right)^{2}}{2^{2}}}$$

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

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}}{2^{2}}-\frac{3^{2}\left(\sqrt{3}\right)^{2}}{2^{2}}}$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}}{2^{2}}-\frac{9\left(\sqrt{3}\right)^{2}}{2^{2}}}$$

The square of \sqrt{3} is 3.

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}}{2^{2}}-\frac{9\times 3}{2^{2}}}$$

Multiply 9 and 3 to get 27.

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}}{2^{2}}-\frac{27}{2^{2}}}$$

Calculate 2 to the power of 2 and get 4.

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}}{2^{2}}-\frac{27}{4}}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}}{4}-\frac{27}{4}}$$

Since \frac{\left(9\sqrt{3}\right)^{2}}{4} and \frac{27}{4} have the same denominator, subtract them by subtracting their numerators.

$$\sqrt{\frac{\left(9\sqrt{3}\right)^{2}-27}{4}}$$

Expand \left(9\sqrt{3}\right)^{2}.

$$\sqrt{\frac{9^{2}\left(\sqrt{3}\right)^{2}-27}{4}}$$

Calculate 9 to the power of 2 and get 81.

$$\sqrt{\frac{81\left(\sqrt{3}\right)^{2}-27}{4}}$$

The square of \sqrt{3} is 3.

$$\sqrt{\frac{81\times 3-27}{4}}$$

Multiply 81 and 3 to get 243.

$$\sqrt{\frac{243-27}{4}}$$

Subtract 27 from 243 to get 216.

$$\sqrt{\frac{216}{4}}$$

Divide 216 by 4 to get 54.

$$\sqrt{54}$$

Factor 54=3^{2}\times 6. Rewrite the square root of the product \sqrt{3^{2}\times 6} as the product of square roots \sqrt{3^{2}}\sqrt{6}. Take the square root of 3^{2}.

$$3\sqrt{6}$$