Solve (sqrt{6}-sqrt{20})^2

$$( \sqrt { 6 } - \sqrt { 20 } ) ^ { 2 }$$

Evaluate

$$26-4\sqrt{30}\approx 4.0910977$$

Solution Steps

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-2\sqrt{5}\right)^{2}.

$$\left(\sqrt{6}\right)^{2}-4\sqrt{6}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{6} is 6.

$$6-4\sqrt{6}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{5}, multiply the numbers under the square root.

$$6-4\sqrt{30}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$6-4\sqrt{30}+4\times 5$$

Multiply 4 and 5 to get 20.

$$6-4\sqrt{30}+20$$

Add 6 and 20 to get 26.

$$26-4\sqrt{30}$$

Expand

$$26-4\sqrt{30}$$

Solution Steps

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-2\sqrt{5}\right)^{2}.

$$\left(\sqrt{6}\right)^{2}-4\sqrt{6}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{6} is 6.

$$6-4\sqrt{6}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{5}, multiply the numbers under the square root.

$$6-4\sqrt{30}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$6-4\sqrt{30}+4\times 5$$

Multiply 4 and 5 to get 20.

$$6-4\sqrt{30}+20$$

Add 6 and 20 to get 26.

$$26-4\sqrt{30}$$