Solve 2/sqrt{8+sqrt{7}}=

$$\frac { 2 } { \sqrt { 8 } + \sqrt { 7 } } =$$

Evaluate

$$2^{\frac{5}{2}}-2\sqrt{7}\approx 0.365351627$$

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}.

$$\frac{2}{2\sqrt{2}+\sqrt{7}}$$

Rationalize the denominator of \frac{2}{2\sqrt{2}+\sqrt{7}} by multiplying numerator and denominator by 2\sqrt{2}-\sqrt{7}.

$$\frac{2\left(2\sqrt{2}-\sqrt{7}\right)}{\left(2\sqrt{2}+\sqrt{7}\right)\left(2\sqrt{2}-\sqrt{7}\right)}$$

Consider \left(2\sqrt{2}+\sqrt{7}\right)\left(2\sqrt{2}-\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{2\left(2\sqrt{2}-\sqrt{7}\right)}{\left(2\sqrt{2}\right)^{2}-\left(\sqrt{7}\right)^{2}}$$

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

$$\frac{2\left(2\sqrt{2}-\sqrt{7}\right)}{2^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{7}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2\left(2\sqrt{2}-\sqrt{7}\right)}{4\left(\sqrt{2}\right)^{2}-\left(\sqrt{7}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{2\left(2\sqrt{2}-\sqrt{7}\right)}{4\times 2-\left(\sqrt{7}\right)^{2}}$$

Multiply 4 and 2 to get 8.

$$\frac{2\left(2\sqrt{2}-\sqrt{7}\right)}{8-\left(\sqrt{7}\right)^{2}}$$

The square of \sqrt{7} is 7.

$$\frac{2\left(2\sqrt{2}-\sqrt{7}\right)}{8-7}$$

Subtract 7 from 8 to get 1.

$$\frac{2\left(2\sqrt{2}-\sqrt{7}\right)}{1}$$

Anything divided by one gives itself.

$$2\left(2\sqrt{2}-\sqrt{7}\right)$$

Use the distributive property to multiply 2 by 2\sqrt{2}-\sqrt{7}.

$$4\sqrt{2}-2\sqrt{7}$$