Solve frac{sqrt{6}-sqrt{2}}{4}divfrac{sqrt{6}+sqrt{2}}{4}

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

Evaluate

$$2-\sqrt{3}\approx 0.267949192$$

Solution Steps

Divide \frac{\sqrt{6}-\sqrt{2}}{4} by \frac{\sqrt{6}+\sqrt{2}}{4} by multiplying \frac{\sqrt{6}-\sqrt{2}}{4} by the reciprocal of \frac{\sqrt{6}+\sqrt{2}}{4}.

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

Cancel out 4 in both numerator and denominator.

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

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

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

Consider \left(\sqrt{2}+\sqrt{6}\right)\left(\sqrt{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}.

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

Square \sqrt{2}. Square \sqrt{6}.

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

Subtract 6 from 2 to get -4.

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

Apply the distributive property by multiplying each term of \sqrt{6}-\sqrt{2} by each term of \sqrt{2}-\sqrt{6}.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{\sqrt{2}\sqrt{3}\sqrt{2}-\left(\sqrt{6}\right)^{2}-\left(\sqrt{2}\right)^{2}+\sqrt{2}\sqrt{6}}{-4}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{2\sqrt{3}-\left(\sqrt{6}\right)^{2}-\left(\sqrt{2}\right)^{2}+\sqrt{2}\sqrt{6}}{-4}$$

The square of \sqrt{6} is 6.

$$\frac{2\sqrt{3}-6-\left(\sqrt{2}\right)^{2}+\sqrt{2}\sqrt{6}}{-4}$$

The square of \sqrt{2} is 2.

$$\frac{2\sqrt{3}-6-2+\sqrt{2}\sqrt{6}}{-4}$$

Subtract 2 from -6 to get -8.

$$\frac{2\sqrt{3}-8+\sqrt{2}\sqrt{6}}{-4}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{2\sqrt{3}-8+\sqrt{2}\sqrt{2}\sqrt{3}}{-4}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{2\sqrt{3}-8+2\sqrt{3}}{-4}$$

Combine 2\sqrt{3} and 2\sqrt{3} to get 4\sqrt{3}.

$$\frac{4\sqrt{3}-8}{-4}$$

Divide each term of 4\sqrt{3}-8 by -4 to get -\sqrt{3}+2.

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