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

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

Evaluate

$$\sqrt{3}+2\sqrt{2}\approx 4.560477932$$

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}.

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

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

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

Consider \left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 8 to get 5.

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

Cancel out 5 and 5.

$$2\sqrt{2}+\sqrt{3}$$