Solve 1/2+sqrt{6}-1/2-sqrt{6}=

$$\frac { 1 } { 2 + \sqrt { 6 } } - \frac { 1 } { 2 - \sqrt { 6 } } =$$

Evaluate

$$\sqrt{6}\approx 2.449489743$$

Short Solution Steps

Rationalize the denominator of \frac{1}{2+\sqrt{6}} by multiplying numerator and denominator by 2-\sqrt{6}.

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

Consider \left(2+\sqrt{6}\right)\left(2-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 2. Square \sqrt{6}.

$$\frac{2-\sqrt{6}}{4-6}-\frac{1}{2-\sqrt{6}}$$

Subtract 6 from 4 to get -2.

$$\frac{2-\sqrt{6}}{-2}-\frac{1}{2-\sqrt{6}}$$

Multiply both numerator and denominator by -1.

$$\frac{-2+\sqrt{6}}{2}-\frac{1}{2-\sqrt{6}}$$

Rationalize the denominator of \frac{1}{2-\sqrt{6}} by multiplying numerator and denominator by 2+\sqrt{6}.

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

Consider \left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 2. Square \sqrt{6}.

$$\frac{-2+\sqrt{6}}{2}-\frac{2+\sqrt{6}}{4-6}$$

Subtract 6 from 4 to get -2.

$$\frac{-2+\sqrt{6}}{2}-\frac{2+\sqrt{6}}{-2}$$

Multiply both numerator and denominator by -1.

$$\frac{-2+\sqrt{6}}{2}-\frac{-2-\sqrt{6}}{2}$$

Since \frac{-2+\sqrt{6}}{2} and \frac{-2-\sqrt{6}}{2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in -2+\sqrt{6}-\left(-2-\sqrt{6}\right).

$$\frac{-2+\sqrt{6}+2+\sqrt{6}}{2}$$

Do the calculations in -2+\sqrt{6}+2+\sqrt{6}.

$$\frac{2\sqrt{6}}{2}$$

Cancel out 2 and 2.

$$\sqrt{6}$$