Solve 3/sqrt{8-sqrt{5}}

$$\frac{ 3 }{ \sqrt{ 8 } - \sqrt{ 5 } }$$

Evaluate

$$\sqrt{5}+2\sqrt{2}\approx 5.064495102$$

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}.

$$\frac{3}{2\sqrt{2}-\sqrt{5}}$$

Rationalize the denominator of \frac{3}{2\sqrt{2}-\sqrt{5}} by multiplying numerator and denominator by 2\sqrt{2}+\sqrt{5}.

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

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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$\frac{3\left(2\sqrt{2}+\sqrt{5}\right)}{4\left(\sqrt{2}\right)^{2}-\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{3\left(2\sqrt{2}+\sqrt{5}\right)}{4\times 2-\left(\sqrt{5}\right)^{2}}$$

Multiply 4 and 2 to get 8.

$$\frac{3\left(2\sqrt{2}+\sqrt{5}\right)}{8-\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{3\left(2\sqrt{2}+\sqrt{5}\right)}{8-5}$$

Subtract 5 from 8 to get 3.

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

Cancel out 3 and 3.

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