Solve frac{2sqrt{2}-sqrt{6}}{2sqrt{2}+sqrt{6}}

$$\frac { 2 \sqrt { 2 } - \sqrt { 6 } } { 2 \sqrt { 2 } + \sqrt { 6 } }$$

Evaluate

$$7-4\sqrt{3}\approx 0.07179677$$

Solution Steps

Rationalize the denominator of \frac{2\sqrt{2}-\sqrt{6}}{2\sqrt{2}+\sqrt{6}} by multiplying numerator and denominator by 2\sqrt{2}-\sqrt{6}.

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

Consider \left(2\sqrt{2}+\sqrt{6}\right)\left(2\sqrt{2}-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Multiply 2\sqrt{2}-\sqrt{6} and 2\sqrt{2}-\sqrt{6} to get \left(2\sqrt{2}-\sqrt{6}\right)^{2}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

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

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}.

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

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

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

Multiply -4 and 2 to get -8.

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

The square of \sqrt{6} is 6.

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

Add 8 and 6 to get 14.

$$\frac{14-8\sqrt{3}}{\left(2\sqrt{2}\right)^{2}-\left(\sqrt{6}\right)^{2}}$$

Expand \left(2\sqrt{2}\right)^{2}.

$$\frac{14-8\sqrt{3}}{2^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{6}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{14-8\sqrt{3}}{4\left(\sqrt{2}\right)^{2}-\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{14-8\sqrt{3}}{4\times 2-\left(\sqrt{6}\right)^{2}}$$

Multiply 4 and 2 to get 8.

$$\frac{14-8\sqrt{3}}{8-\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{14-8\sqrt{3}}{8-6}$$

Subtract 6 from 8 to get 2.

$$\frac{14-8\sqrt{3}}{2}$$

Divide each term of 14-8\sqrt{3} by 2 to get 7-4\sqrt{3}.

$$7-4\sqrt{3}$$