Solve sqrt{(2sqrt{3}-2)^2+16+(2sqrt{3}-2)*4}

$$\sqrt { ( 2 \sqrt { 3 } - 2 ) ^ { 2 } + 16 + ( 2 \sqrt { 3 } - 2 ) \times 4 }$$

Evaluate

$$2\sqrt{6}\approx 4.898979486$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Add 12 and 4 to get 16.

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

Add 16 and 16 to get 32.

$$\sqrt{32-8\sqrt{3}+4\left(2\sqrt{3}-2\right)}$$

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

$$\sqrt{32-8\sqrt{3}+8\sqrt{3}-8}$$

Combine -8\sqrt{3} and 8\sqrt{3} to get 0.

$$\sqrt{32-8}$$

Subtract 8 from 32 to get 24.

$$\sqrt{24}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$2\sqrt{6}$$