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

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

Evaluate

$$\frac{7\sqrt{3}}{3}-4\approx 0.041451884$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

$$\frac{3-4\sqrt{3}+4}{\sqrt{3}}$$

Add 3 and 4 to get 7.

$$\frac{7-4\sqrt{3}}{\sqrt{3}}$$

Rationalize the denominator of \frac{7-4\sqrt{3}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(7-4\sqrt{3}\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(7-4\sqrt{3}\right)\sqrt{3}}{3}$$

Use the distributive property to multiply 7-4\sqrt{3} by \sqrt{3}.

$$\frac{7\sqrt{3}-4\left(\sqrt{3}\right)^{2}}{3}$$

The square of \sqrt{3} is 3.

$$\frac{7\sqrt{3}-4\times 3}{3}$$

Multiply -4 and 3 to get -12.

$$\frac{7\sqrt{3}-12}{3}$$