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

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

Evaluate

$$12-4\sqrt{3}\approx 5.07179677$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

$$4\times 3-4\sqrt{3}+1+\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)$$

Multiply 4 and 3 to get 12.

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

Add 12 and 1 to get 13.

$$13-4\sqrt{3}+\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)$$

Consider \left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$13-4\sqrt{3}+\left(\sqrt{3}\right)^{2}-4$$

The square of \sqrt{3} is 3.

$$13-4\sqrt{3}+3-4$$

Subtract 4 from 3 to get -1.

$$13-4\sqrt{3}-1$$

Subtract 1 from 13 to get 12.

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