Solve (sqrt{3}-2)(sqrt{3}-2)+sqrt{12}+6sqrt{1/3}

$$( \sqrt{ 3 } -2)( \sqrt{ 3 } -2)+ \sqrt{ 12 } +6 \sqrt{ \frac{ 1 }{ 3 } }$$

Evaluate

$$7$$

Solution Steps

Multiply \sqrt{3}-2 and \sqrt{3}-2 to get \left(\sqrt{3}-2\right)^{2}.

$$\left(\sqrt{3}-2\right)^{2}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

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

$$\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

The square of \sqrt{3} is 3.

$$3-4\sqrt{3}+4+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Add 3 and 4 to get 7.

$$7-4\sqrt{3}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$7-4\sqrt{3}+2\sqrt{3}+6\sqrt{\frac{1}{3}}$$

Combine -4\sqrt{3} and 2\sqrt{3} to get -2\sqrt{3}.

$$7-2\sqrt{3}+6\sqrt{\frac{1}{3}}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$7-2\sqrt{3}+6\times \frac{\sqrt{1}}{\sqrt{3}}$$

Calculate the square root of 1 and get 1.

$$7-2\sqrt{3}+6\times \frac{1}{\sqrt{3}}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

$$7-2\sqrt{3}+6\times \frac{\sqrt{3}}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$7-2\sqrt{3}+2\sqrt{3}$$

Combine -2\sqrt{3} and 2\sqrt{3} to get 0.

$$7$$

Factor

$$7$$