Solve 6/2sqrt{2+sqrt{5}}

$$\frac { 6 } { 2 \sqrt { 2 } + \sqrt { 5 } }$$

Evaluate

$$2^{\frac{5}{2}}-2\sqrt{5}\approx 1.184718294$$

Solution Steps

Rationalize the denominator of \frac{6}{2\sqrt{2}+\sqrt{5}} by multiplying numerator and denominator by 2\sqrt{2}-\sqrt{5}.

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

Consider \left(2\sqrt{2}+\sqrt{5}\right)\left(2\sqrt{2}-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{6\left(2\sqrt{2}-\sqrt{5}\right)}{8-\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{6\left(2\sqrt{2}-\sqrt{5}\right)}{8-5}$$

Subtract 5 from 8 to get 3.

$$\frac{6\left(2\sqrt{2}-\sqrt{5}\right)}{3}$$

Divide 6\left(2\sqrt{2}-\sqrt{5}\right) by 3 to get 2\left(2\sqrt{2}-\sqrt{5}\right).

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

Use the distributive property to multiply 2 by 2\sqrt{2}-\sqrt{5}.

$$4\sqrt{2}-2\sqrt{5}$$