Solve {left(sqrt{6}+sqrt{2}right)}^2

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

Evaluate

$$4\left(\sqrt{3}+2\right)\approx 14.92820323$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply 2 and 2 to get 4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

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

Expand

$$4 \sqrt{3} + 8 = 14.92820323$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply 2 and 2 to get 4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

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