Solve (3sqrt{5}-sqrt{2})^2

$$( 3 \sqrt { 5 } - \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$47-6\sqrt{10}\approx 28.026334039$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-\sqrt{2}\right)^{2}.

$$9\left(\sqrt{5}\right)^{2}-6\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9\times 5-6\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

Multiply 9 and 5 to get 45.

$$45-6\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$45-6\sqrt{10}+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$45-6\sqrt{10}+2$$

Add 45 and 2 to get 47.

$$47-6\sqrt{10}$$

Expand

$$47-6\sqrt{10}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-\sqrt{2}\right)^{2}.

$$9\left(\sqrt{5}\right)^{2}-6\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9\times 5-6\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

Multiply 9 and 5 to get 45.

$$45-6\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$45-6\sqrt{10}+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$45-6\sqrt{10}+2$$

Add 45 and 2 to get 47.

$$47-6\sqrt{10}$$