Solve 4+(sqrt{6}-sqrt{2})^2

$$4 + ( \sqrt { 6 } - \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$12-4\sqrt{3}\approx 5.07179677$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

$$4+6-4\sqrt{3}+2$$

Add 6 and 2 to get 8.

$$4+8-4\sqrt{3}$$

Add 4 and 8 to get 12.

$$12-4\sqrt{3}$$

Expand

$$12-4\sqrt{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

$$4+6-4\sqrt{3}+2$$

Add 6 and 2 to get 8.

$$4+8-4\sqrt{3}$$

Add 4 and 8 to get 12.

$$12-4\sqrt{3}$$