Solve {left(sqrt{2}+sqrt{6}right)}^2-(5-3)

$${ \left( \sqrt{ 2 } + \sqrt{ 6 } \right) }^{ 2 } -(5-3)$$

Evaluate

$$4\sqrt{3}+6\approx 12.92820323$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

The square of \sqrt{2} is 2.

$$2+2\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

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}.

$$2+2\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

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

$$2+2\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

Multiply 2 and 2 to get 4.

$$2+4\sqrt{3}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

The square of \sqrt{6} is 6.

$$2+4\sqrt{3}+6-\left(5-3\right)$$

Add 2 and 6 to get 8.

$$8+4\sqrt{3}-\left(5-3\right)$$

Subtract 3 from 5 to get 2.

$$8+4\sqrt{3}-2$$

Subtract 2 from 8 to get 6.

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

Expand

$$4 \sqrt{3} + 6 = 12.92820323$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

The square of \sqrt{2} is 2.

$$2+2\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

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}.

$$2+2\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

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

$$2+2\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

Multiply 2 and 2 to get 4.

$$2+4\sqrt{3}+\left(\sqrt{6}\right)^{2}-\left(5-3\right)$$

The square of \sqrt{6} is 6.

$$2+4\sqrt{3}+6-\left(5-3\right)$$

Add 2 and 6 to get 8.

$$8+4\sqrt{3}-\left(5-3\right)$$

Subtract 3 from 5 to get 2.

$$8+4\sqrt{3}-2$$

Subtract 2 from 8 to get 6.

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