Solve (2sqrt{5}-sqrt{15})^2

$$( 2 \sqrt { 5 } - \sqrt { 15 } ) ^ { 2 }$$

Evaluate

$$35-20\sqrt{3}\approx 0.358983849$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{5}-\sqrt{15}\right)^{2}.

$$4\left(\sqrt{5}\right)^{2}-4\sqrt{5}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4\times 5-4\sqrt{5}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

Multiply 4 and 5 to get 20.

$$20-4\sqrt{5}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$20-4\sqrt{5}\sqrt{5}\sqrt{3}+\left(\sqrt{15}\right)^{2}$$

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

$$20-4\times 5\sqrt{3}+\left(\sqrt{15}\right)^{2}$$

Multiply -4 and 5 to get -20.

$$20-20\sqrt{3}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{15} is 15.

$$20-20\sqrt{3}+15$$

Add 20 and 15 to get 35.

$$35-20\sqrt{3}$$

Expand

$$35-20\sqrt{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{5}-\sqrt{15}\right)^{2}.

$$4\left(\sqrt{5}\right)^{2}-4\sqrt{5}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4\times 5-4\sqrt{5}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

Multiply 4 and 5 to get 20.

$$20-4\sqrt{5}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$20-4\sqrt{5}\sqrt{5}\sqrt{3}+\left(\sqrt{15}\right)^{2}$$

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

$$20-4\times 5\sqrt{3}+\left(\sqrt{15}\right)^{2}$$

Multiply -4 and 5 to get -20.

$$20-20\sqrt{3}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{15} is 15.

$$20-20\sqrt{3}+15$$

Add 20 and 15 to get 35.

$$35-20\sqrt{3}$$