Solve 8+sqrt{3}+4sqrt{3}*(2-sqrt{3})^2

$$8 + \sqrt { 3 } + 4 \sqrt { 3 } \times ( 2 - \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$29\sqrt{3}-40\approx 10.229473419$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

$$8+\sqrt{3}+4\sqrt{3}\left(4-4\sqrt{3}+3\right)$$

Add 4 and 3 to get 7.

$$8+\sqrt{3}+4\sqrt{3}\left(7-4\sqrt{3}\right)$$

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

$$8+\sqrt{3}+28\sqrt{3}-16\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$8+\sqrt{3}+28\sqrt{3}-16\times 3$$

Multiply -16 and 3 to get -48.

$$8+\sqrt{3}+28\sqrt{3}-48$$

Combine \sqrt{3} and 28\sqrt{3} to get 29\sqrt{3}.

$$8+29\sqrt{3}-48$$

Subtract 48 from 8 to get -40.

$$-40+29\sqrt{3}$$