Solve frac{sqrt{7}-1}{sqrt{7+1}}+frac{sqrt{7}+1}{sqrt{7}-1}

$$\frac{ \sqrt{ 7 } -1 }{ \sqrt{ 7+1 } } + \frac{ \sqrt{ 7 } +1 }{ \sqrt{ 7 } -1 }$$

Evaluate

$$\frac{\sqrt{7}}{3}+\frac{\sqrt{14}}{4}-\frac{\sqrt{2}}{4}+\frac{4}{3}\approx 2.797111393$$

Short Solution Steps

Add 7 and 1 to get 8.

$$\frac{\sqrt{7}-1}{\sqrt{8}}+\frac{\sqrt{7}+1}{\sqrt{7}-1}$$

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}.

$$\frac{\sqrt{7}-1}{2\sqrt{2}}+\frac{\sqrt{7}+1}{\sqrt{7}-1}$$

Rationalize the denominator of \frac{\sqrt{7}-1}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{\sqrt{7}+1}{\sqrt{7}-1}$$

Rationalize the denominator of \frac{\sqrt{7}+1}{\sqrt{7}-1} by multiplying numerator and denominator by \sqrt{7}+1.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{\left(\sqrt{7}+1\right)\left(\sqrt{7}+1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}$$

Consider \left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{\left(\sqrt{7}+1\right)\left(\sqrt{7}+1\right)}{\left(\sqrt{7}\right)^{2}-1^{2}}$$

Square \sqrt{7}. Square 1.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{\left(\sqrt{7}+1\right)\left(\sqrt{7}+1\right)}{7-1}$$

Subtract 1 from 7 to get 6.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{\left(\sqrt{7}+1\right)\left(\sqrt{7}+1\right)}{6}$$

Multiply \sqrt{7}+1 and \sqrt{7}+1 to get \left(\sqrt{7}+1\right)^{2}.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{\left(\sqrt{7}+1\right)^{2}}{6}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+1\right)^{2}.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{\left(\sqrt{7}\right)^{2}+2\sqrt{7}+1}{6}$$

The square of \sqrt{7} is 7.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{7+2\sqrt{7}+1}{6}$$

Add 7 and 1 to get 8.

$$\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4}+\frac{8+2\sqrt{7}}{6}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 6 is 12. Multiply \frac{\left(\sqrt{7}-1\right)\sqrt{2}}{4} times \frac{3}{3}. Multiply \frac{8+2\sqrt{7}}{6} times \frac{2}{2}.

$$\frac{3\left(\sqrt{7}-1\right)\sqrt{2}}{12}+\frac{2\left(8+2\sqrt{7}\right)}{12}$$

Since \frac{3\left(\sqrt{7}-1\right)\sqrt{2}}{12} and \frac{2\left(8+2\sqrt{7}\right)}{12} have the same denominator, add them by adding their numerators.

$$\frac{3\left(\sqrt{7}-1\right)\sqrt{2}+2\left(8+2\sqrt{7}\right)}{12}$$

Do the multiplications in 3\left(\sqrt{7}-1\right)\sqrt{2}+2\left(8+2\sqrt{7}\right).

$$\frac{3\sqrt{14}-3\sqrt{2}+16+4\sqrt{7}}{12}$$

Factor

$$\frac{3 \sqrt{14} + 4 \sqrt{7} + 16 - 3 \sqrt{2}}{12} = 2.797111393121742$$