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

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

Evaluate

$$13-6\sqrt{3}\approx 2.607695155$$

Solution Steps

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

$$\frac{1}{2}\left(3+4\left(\sqrt{3}\right)^{2}-8\sqrt{3}+4+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

The square of \sqrt{3} is 3.

$$\frac{1}{2}\left(3+4\times 3-8\sqrt{3}+4+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

Multiply 4 and 3 to get 12.

$$\frac{1}{2}\left(3+12-8\sqrt{3}+4+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

Add 12 and 4 to get 16.

$$\frac{1}{2}\left(3+16-8\sqrt{3}+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

Add 3 and 16 to get 19.

$$\frac{1}{2}\left(19-8\sqrt{3}+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

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

$$\frac{1}{2}\left(19-8\sqrt{3}+\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4\right)^{1}$$

The square of \sqrt{3} is 3.

$$\frac{1}{2}\left(19-8\sqrt{3}+3-4\sqrt{3}+4\right)^{1}$$

Add 3 and 4 to get 7.

$$\frac{1}{2}\left(19-8\sqrt{3}+7-4\sqrt{3}\right)^{1}$$

Add 19 and 7 to get 26.

$$\frac{1}{2}\left(26-8\sqrt{3}-4\sqrt{3}\right)^{1}$$

Combine -8\sqrt{3} and -4\sqrt{3} to get -12\sqrt{3}.

$$\frac{1}{2}\left(26-12\sqrt{3}\right)^{1}$$

Calculate 26-12\sqrt{3} to the power of 1 and get 26-12\sqrt{3}.

$$\frac{1}{2}\left(26-12\sqrt{3}\right)$$

Use the distributive property to multiply \frac{1}{2} by 26-12\sqrt{3}.

$$13-6\sqrt{3}$$

Expand

$$13-6\sqrt{3}$$

Solution Steps

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

$$\frac{1}{2}\left(3+4\left(\sqrt{3}\right)^{2}-8\sqrt{3}+4+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

The square of \sqrt{3} is 3.

$$\frac{1}{2}\left(3+4\times 3-8\sqrt{3}+4+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

Multiply 4 and 3 to get 12.

$$\frac{1}{2}\left(3+12-8\sqrt{3}+4+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

Add 12 and 4 to get 16.

$$\frac{1}{2}\left(3+16-8\sqrt{3}+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

Add 3 and 16 to get 19.

$$\frac{1}{2}\left(19-8\sqrt{3}+\left(\sqrt{3}-2\right)^{2}\right)^{1}$$

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

$$\frac{1}{2}\left(19-8\sqrt{3}+\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4\right)^{1}$$

The square of \sqrt{3} is 3.

$$\frac{1}{2}\left(19-8\sqrt{3}+3-4\sqrt{3}+4\right)^{1}$$

Add 3 and 4 to get 7.

$$\frac{1}{2}\left(19-8\sqrt{3}+7-4\sqrt{3}\right)^{1}$$

Add 19 and 7 to get 26.

$$\frac{1}{2}\left(26-8\sqrt{3}-4\sqrt{3}\right)^{1}$$

Combine -8\sqrt{3} and -4\sqrt{3} to get -12\sqrt{3}.

$$\frac{1}{2}\left(26-12\sqrt{3}\right)^{1}$$

Calculate 26-12\sqrt{3} to the power of 1 and get 26-12\sqrt{3}.

$$\frac{1}{2}\left(26-12\sqrt{3}\right)$$

Use the distributive property to multiply \frac{1}{2} by 26-12\sqrt{3}.

$$13-6\sqrt{3}$$