Solve sqrt{(2sqrt{3}-4)^2+8(2sqrt{3}-4)+1}

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

Evaluate

$$\text{Indeterminate}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-4\right)^{2}.

$$\sqrt{4\left(\sqrt{3}\right)^{2}-16\sqrt{3}+16+8\left(2\sqrt{3}-4\right)+1}$$

The square of \sqrt{3} is 3.

$$\sqrt{4\times 3-16\sqrt{3}+16+8\left(2\sqrt{3}-4\right)+1}$$

Multiply 4 and 3 to get 12.

$$\sqrt{12-16\sqrt{3}+16+8\left(2\sqrt{3}-4\right)+1}$$

Add 12 and 16 to get 28.

$$\sqrt{28-16\sqrt{3}+8\left(2\sqrt{3}-4\right)+1}$$

Use the distributive property to multiply 8 by 2\sqrt{3}-4.

$$\sqrt{28-16\sqrt{3}+16\sqrt{3}-32+1}$$

Combine -16\sqrt{3} and 16\sqrt{3} to get 0.

$$\sqrt{28-32+1}$$

Subtract 32 from 28 to get -4.

$$\sqrt{-4+1}$$

Add -4 and 1 to get -3.

$$\sqrt{-3}$$