Solve 16/sqrt{5-sqrt{8}}

$$\frac { 16 } { \sqrt { 5 } - \sqrt { 8 } }$$

Evaluate

$$\frac{-16\sqrt{5}-32\sqrt{2}}{3}\approx -27.010640545$$

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}.

$$\frac{16}{\sqrt{5}-2\sqrt{2}}$$

Rationalize the denominator of \frac{16}{\sqrt{5}-2\sqrt{2}} by multiplying numerator and denominator by \sqrt{5}+2\sqrt{2}.

$$\frac{16\left(\sqrt{5}+2\sqrt{2}\right)}{\left(\sqrt{5}-2\sqrt{2}\right)\left(\sqrt{5}+2\sqrt{2}\right)}$$

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

$$\frac{16\left(\sqrt{5}+2\sqrt{2}\right)}{\left(\sqrt{5}\right)^{2}-\left(-2\sqrt{2}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{16\left(\sqrt{5}+2\sqrt{2}\right)}{5-\left(-2\sqrt{2}\right)^{2}}$$

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

$$\frac{16\left(\sqrt{5}+2\sqrt{2}\right)}{5-\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{16\left(\sqrt{5}+2\sqrt{2}\right)}{5-4\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{16\left(\sqrt{5}+2\sqrt{2}\right)}{5-4\times 2}$$

Multiply 4 and 2 to get 8.

$$\frac{16\left(\sqrt{5}+2\sqrt{2}\right)}{5-8}$$

Subtract 8 from 5 to get -3.

$$\frac{16\left(\sqrt{5}+2\sqrt{2}\right)}{-3}$$

Use the distributive property to multiply 16 by \sqrt{5}+2\sqrt{2}.

$$\frac{16\sqrt{5}+32\sqrt{2}}{-3}$$