Solve {left(sqrt{2}+sqrt{10}right)}^2

$${ \left( \sqrt{ 2 } + \sqrt{ 10 } \right) }^{ 2 }$$

Evaluate

$$4\left(\sqrt{5}+3\right)\approx 20.94427191$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+\sqrt{10}\right)^{2}.

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{2} is 2.

$$2+2\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

$$2+2\sqrt{2}\sqrt{2}\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

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

$$2+2\times 2\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$2+4\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$2+4\sqrt{5}+10$$

Add 2 and 10 to get 12.

$$12+4\sqrt{5}$$

Expand

$$4 \sqrt{5} + 12 = 20.94427191$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+\sqrt{10}\right)^{2}.

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{2} is 2.

$$2+2\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

$$2+2\sqrt{2}\sqrt{2}\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

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

$$2+2\times 2\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$2+4\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$2+4\sqrt{5}+10$$

Add 2 and 10 to get 12.

$$12+4\sqrt{5}$$