Solve (3sqrt{2}-4sqrt{5})^2

$$( 3 \sqrt { 2 } - 4 \sqrt { 5 } ) ^ { 2 }$$

Evaluate

$$98-24\sqrt{10}\approx 22.105336156$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{2}-4\sqrt{5}\right)^{2}.

$$9\left(\sqrt{2}\right)^{2}-24\sqrt{2}\sqrt{5}+16\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9\times 2-24\sqrt{2}\sqrt{5}+16\left(\sqrt{5}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$18-24\sqrt{2}\sqrt{5}+16\left(\sqrt{5}\right)^{2}$$

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

$$18-24\sqrt{10}+16\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$18-24\sqrt{10}+16\times 5$$

Multiply 16 and 5 to get 80.

$$18-24\sqrt{10}+80$$

Add 18 and 80 to get 98.

$$98-24\sqrt{10}$$

Expand

$$98-24\sqrt{10}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{2}-4\sqrt{5}\right)^{2}.

$$9\left(\sqrt{2}\right)^{2}-24\sqrt{2}\sqrt{5}+16\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9\times 2-24\sqrt{2}\sqrt{5}+16\left(\sqrt{5}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$18-24\sqrt{2}\sqrt{5}+16\left(\sqrt{5}\right)^{2}$$

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

$$18-24\sqrt{10}+16\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$18-24\sqrt{10}+16\times 5$$

Multiply 16 and 5 to get 80.

$$18-24\sqrt{10}+80$$

Add 18 and 80 to get 98.

$$98-24\sqrt{10}$$