Solve {left(3sqrt{2}-sqrt{10}right)}^2

$${ \left(3 \sqrt{ 2 } - \sqrt{ 10 } \right) }^{ 2 }$$

Evaluate

$$28-12\sqrt{5}\approx 1.16718427$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{2}-\sqrt{10}\right)^{2}.

$$9\left(\sqrt{2}\right)^{2}-6\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9\times 2-6\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$18-6\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

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}.

$$18-6\sqrt{2}\sqrt{2}\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

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

$$18-6\times 2\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

Multiply -6 and 2 to get -12.

$$18-12\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$18-12\sqrt{5}+10$$

Add 18 and 10 to get 28.

$$28-12\sqrt{5}$$

Expand

$$28-12\sqrt{5}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{2}-\sqrt{10}\right)^{2}.

$$9\left(\sqrt{2}\right)^{2}-6\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9\times 2-6\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$18-6\sqrt{2}\sqrt{10}+\left(\sqrt{10}\right)^{2}$$

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}.

$$18-6\sqrt{2}\sqrt{2}\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

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

$$18-6\times 2\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

Multiply -6 and 2 to get -12.

$$18-12\sqrt{5}+\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$18-12\sqrt{5}+10$$

Add 18 and 10 to get 28.

$$28-12\sqrt{5}$$