Solve 1/sqrt{3+1}+1/sqrt{3-1}

$$\frac{ 1 }{ \sqrt{ 3 } +1 } + \frac{ 1 }{ \sqrt{ 3 } -1 }$$

Evaluate

$$\sqrt{3}\approx 1.732050808$$

Short Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{3}+1} by multiplying numerator and denominator by \sqrt{3}-1.

$$\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{1}{\sqrt{3}-1}$$

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-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}.

$$\frac{\sqrt{3}-1}{\left(\sqrt{3}\right)^{2}-1^{2}}+\frac{1}{\sqrt{3}-1}$$

Square \sqrt{3}. Square 1.

$$\frac{\sqrt{3}-1}{3-1}+\frac{1}{\sqrt{3}-1}$$

Subtract 1 from 3 to get 2.

$$\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{3}-1}$$

Rationalize the denominator of \frac{1}{\sqrt{3}-1} by multiplying numerator and denominator by \sqrt{3}+1.

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

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+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}.

$$\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{\left(\sqrt{3}\right)^{2}-1^{2}}$$

Square \sqrt{3}. Square 1.

$$\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{3-1}$$

Subtract 1 from 3 to get 2.

$$\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2}$$

Since \frac{\sqrt{3}-1}{2} and \frac{\sqrt{3}+1}{2} have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{3}-1+\sqrt{3}+1}{2}$$

Do the calculations in \sqrt{3}-1+\sqrt{3}+1.

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

Cancel out 2 and 2.

$$\sqrt{3}$$