Solve 16/2sqrt{2+2sqrt{6}}

$$\frac{ 16 }{ 2 \sqrt{ 2 } +2 \sqrt{ 6 } }$$

Evaluate

$$2\left(\sqrt{6}-\sqrt{2}\right)\approx 2.070552361$$

Solution Steps

Rationalize the denominator of \frac{16}{2\sqrt{2}+2\sqrt{6}} by multiplying numerator and denominator by 2\sqrt{2}-2\sqrt{6}.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{\left(2\sqrt{2}+2\sqrt{6}\right)\left(2\sqrt{2}-2\sqrt{6}\right)}$$

Consider \left(2\sqrt{2}+2\sqrt{6}\right)\left(2\sqrt{2}-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}.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{\left(2\sqrt{2}\right)^{2}-\left(2\sqrt{6}\right)^{2}}$$

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

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{2^{2}\left(\sqrt{2}\right)^{2}-\left(2\sqrt{6}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{4\left(\sqrt{2}\right)^{2}-\left(2\sqrt{6}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{4\times 2-\left(2\sqrt{6}\right)^{2}}$$

Multiply 4 and 2 to get 8.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{8-\left(2\sqrt{6}\right)^{2}}$$

Expand \left(2\sqrt{6}\right)^{2}.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{8-2^{2}\left(\sqrt{6}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{8-4\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{8-4\times 6}$$

Multiply 4 and 6 to get 24.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{8-24}$$

Subtract 24 from 8 to get -16.

$$\frac{16\left(2\sqrt{2}-2\sqrt{6}\right)}{-16}$$

Cancel out -16 and -16.

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

To find the opposite of 2\sqrt{2}-2\sqrt{6}, find the opposite of each term.

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

The opposite of -2\sqrt{6} is 2\sqrt{6}.

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