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

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

Evaluate

$$10$$

Solution Steps

Calculate the square root of 4 and get 2.

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

Calculate the square root of 4 and get 2.

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

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

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

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

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

Square \sqrt{6}. Square 2.

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

Subtract 4 from 6 to get 2.

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

Multiply \sqrt{6}+2 and \sqrt{6}+2 to get \left(\sqrt{6}+2\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+2\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Add 6 and 4 to get 10.

$$\frac{10+4\sqrt{6}}{2}+\frac{\sqrt{6}-\sqrt{4}}{\sqrt{6}+\sqrt{4}}$$

Divide each term of 10+4\sqrt{6} by 2 to get 5+2\sqrt{6}.

$$5+2\sqrt{6}+\frac{\sqrt{6}-\sqrt{4}}{\sqrt{6}+\sqrt{4}}$$

Calculate the square root of 4 and get 2.

$$5+2\sqrt{6}+\frac{\sqrt{6}-2}{\sqrt{6}+\sqrt{4}}$$

Calculate the square root of 4 and get 2.

$$5+2\sqrt{6}+\frac{\sqrt{6}-2}{\sqrt{6}+2}$$

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

$$5+2\sqrt{6}+\frac{\left(\sqrt{6}-2\right)\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}$$

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

$$5+2\sqrt{6}+\frac{\left(\sqrt{6}-2\right)\left(\sqrt{6}-2\right)}{\left(\sqrt{6}\right)^{2}-2^{2}}$$

Square \sqrt{6}. Square 2.

$$5+2\sqrt{6}+\frac{\left(\sqrt{6}-2\right)\left(\sqrt{6}-2\right)}{6-4}$$

Subtract 4 from 6 to get 2.

$$5+2\sqrt{6}+\frac{\left(\sqrt{6}-2\right)\left(\sqrt{6}-2\right)}{2}$$

Multiply \sqrt{6}-2 and \sqrt{6}-2 to get \left(\sqrt{6}-2\right)^{2}.

$$5+2\sqrt{6}+\frac{\left(\sqrt{6}-2\right)^{2}}{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-2\right)^{2}.

$$5+2\sqrt{6}+\frac{\left(\sqrt{6}\right)^{2}-4\sqrt{6}+4}{2}$$

The square of \sqrt{6} is 6.

$$5+2\sqrt{6}+\frac{6-4\sqrt{6}+4}{2}$$

Add 6 and 4 to get 10.

$$5+2\sqrt{6}+\frac{10-4\sqrt{6}}{2}$$

Divide each term of 10-4\sqrt{6} by 2 to get 5-2\sqrt{6}.

$$5+2\sqrt{6}+5-2\sqrt{6}$$

Add 5 and 5 to get 10.

$$10+2\sqrt{6}-2\sqrt{6}$$

Combine 2\sqrt{6} and -2\sqrt{6} to get 0.

$$10$$

Factor

$$2\times 5$$