Solve (sqrt{10}-sqrt{5})^2=

$$( \sqrt { 10 } - \sqrt { 5 } ) ^ { 2 } =$$

Evaluate

$$15-10\sqrt{2}\approx 0.857864376$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{10}-\sqrt{5}\right)^{2}.

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

The square of \sqrt{10} is 10.

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

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

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

Multiply \sqrt{5} and \sqrt{5} to get 5.

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

Multiply -2 and 5 to get -10.

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

The square of \sqrt{5} is 5.

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

Add 10 and 5 to get 15.

$$15-10\sqrt{2}$$

Expand

$$15-10\sqrt{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{10}-\sqrt{5}\right)^{2}.

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

The square of \sqrt{10} is 10.

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

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

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

Multiply \sqrt{5} and \sqrt{5} to get 5.

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

Multiply -2 and 5 to get -10.

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

The square of \sqrt{5} is 5.

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

Add 10 and 5 to get 15.

$$15-10\sqrt{2}$$