Solve (sqrt{10}-sqrt{2})^2

$$( \sqrt { 10 } - \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$12-4\sqrt{5}\approx 3.05572809$$

Solution Steps

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

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

The square of \sqrt{10} is 10.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 10 and 2 to get 12.

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

Expand

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

Solution Steps

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

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

The square of \sqrt{10} is 10.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 10 and 2 to get 12.

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