Solve (sqrt{8}-sqrt{6})^2=

$$( \sqrt { 8 } - \sqrt { 6 } ) ^ { 2 } =$$

Evaluate

$$14-8\sqrt{3}\approx 0.143593539$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$8-4\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$8-4\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

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

$$8-4\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

Multiply -4 and 2 to get -8.

$$8-8\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$8-8\sqrt{3}+6$$

Add 8 and 6 to get 14.

$$14-8\sqrt{3}$$

Expand

$$14-8\sqrt{3}$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$8-4\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$8-4\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

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

$$8-4\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

Multiply -4 and 2 to get -8.

$$8-8\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$8-8\sqrt{3}+6$$

Add 8 and 6 to get 14.

$$14-8\sqrt{3}$$