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

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

Evaluate

$$4$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

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

The square of \sqrt{3} is 3.

$$3-2\sqrt{3}+1+\sqrt{12}$$

Add 3 and 1 to get 4.

$$4-2\sqrt{3}+\sqrt{12}$$

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}.

$$4-2\sqrt{3}+2\sqrt{3}$$

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

$$4$$

Factor

$$2^{2}$$