Solve (sqrt{12}-2sqrt{2})^2+8sqrt{6}

$$( \sqrt { 12 } - 2 \sqrt { 2 } ) ^ { 2 } + 8 \sqrt { 6 }$$

Evaluate

$$20$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\left(2\sqrt{3}-2\sqrt{2}\right)^{2}+8\sqrt{6}$$

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

$$4\left(\sqrt{3}\right)^{2}-8\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}+8\sqrt{6}$$

The square of \sqrt{3} is 3.

$$4\times 3-8\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}+8\sqrt{6}$$

Multiply 4 and 3 to get 12.

$$12-8\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}+8\sqrt{6}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$12-8\sqrt{6}+4\left(\sqrt{2}\right)^{2}+8\sqrt{6}$$

The square of \sqrt{2} is 2.

$$12-8\sqrt{6}+4\times 2+8\sqrt{6}$$

Multiply 4 and 2 to get 8.

$$12-8\sqrt{6}+8+8\sqrt{6}$$

Add 12 and 8 to get 20.

$$20-8\sqrt{6}+8\sqrt{6}$$

Combine -8\sqrt{6} and 8\sqrt{6} to get 0.

$$20$$

Factor

$$2^{2}\times 5$$