Solve (frac{sqrt{3}-4}{6})^2

$$( \frac { \sqrt { 3 } - 4 } { 6 } ) ^ { 2 }$$

Evaluate

$$-\frac{2\sqrt{3}}{9}+\frac{19}{36}\approx 0.142877598$$

Solution Steps

To raise \frac{\sqrt{3}-4}{6} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-4\right)^{2}.

$$\frac{\left(\sqrt{3}\right)^{2}-8\sqrt{3}+16}{6^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3-8\sqrt{3}+16}{6^{2}}$$

Add 3 and 16 to get 19.

$$\frac{19-8\sqrt{3}}{6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{19-8\sqrt{3}}{36}$$

Expand

$$-\frac{2 \sqrt{3}}{9} + \frac{19}{36} = 0.142877598$$

Solution Steps

To raise \frac{\sqrt{3}-4}{6} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-4\right)^{2}.

$$\frac{\left(\sqrt{3}\right)^{2}-8\sqrt{3}+16}{6^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3-8\sqrt{3}+16}{6^{2}}$$

Add 3 and 16 to get 19.

$$\frac{19-8\sqrt{3}}{6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{19-8\sqrt{3}}{36}$$