Solve frac{sqrt{11}+sqrt{9}}{sqrt{11}-sqrt{9}}+frac{sqrt{11}-sqrt{9}}{sqrt{11}+sqrt{9}}=?

$$\frac { \sqrt { 11 } + \sqrt { 9 } } { \sqrt { 11 } - \sqrt { 9 } } + \frac { \sqrt { 11 } - \sqrt { 9 } } { \sqrt { 11 } + \sqrt { 9 } } = ?$$

Evaluate

$$20$$

Solution Steps

Calculate the square root of 9 and get 3.

$$\frac{\sqrt{11}+3}{\sqrt{11}-\sqrt{9}}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Calculate the square root of 9 and get 3.

$$\frac{\sqrt{11}+3}{\sqrt{11}-3}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Rationalize the denominator of \frac{\sqrt{11}+3}{\sqrt{11}-3} by multiplying numerator and denominator by \sqrt{11}+3.

$$\frac{\left(\sqrt{11}+3\right)\left(\sqrt{11}+3\right)}{\left(\sqrt{11}-3\right)\left(\sqrt{11}+3\right)}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Consider \left(\sqrt{11}-3\right)\left(\sqrt{11}+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{11}+3\right)\left(\sqrt{11}+3\right)}{\left(\sqrt{11}\right)^{2}-3^{2}}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Square \sqrt{11}. Square 3.

$$\frac{\left(\sqrt{11}+3\right)\left(\sqrt{11}+3\right)}{11-9}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Subtract 9 from 11 to get 2.

$$\frac{\left(\sqrt{11}+3\right)\left(\sqrt{11}+3\right)}{2}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Multiply \sqrt{11}+3 and \sqrt{11}+3 to get \left(\sqrt{11}+3\right)^{2}.

$$\frac{\left(\sqrt{11}+3\right)^{2}}{2}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{11}+3\right)^{2}.

$$\frac{\left(\sqrt{11}\right)^{2}+6\sqrt{11}+9}{2}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

The square of \sqrt{11} is 11.

$$\frac{11+6\sqrt{11}+9}{2}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Add 11 and 9 to get 20.

$$\frac{20+6\sqrt{11}}{2}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Divide each term of 20+6\sqrt{11} by 2 to get 10+3\sqrt{11}.

$$10+3\sqrt{11}+\frac{\sqrt{11}-\sqrt{9}}{\sqrt{11}+\sqrt{9}}$$

Calculate the square root of 9 and get 3.

$$10+3\sqrt{11}+\frac{\sqrt{11}-3}{\sqrt{11}+\sqrt{9}}$$

Calculate the square root of 9 and get 3.

$$10+3\sqrt{11}+\frac{\sqrt{11}-3}{\sqrt{11}+3}$$

Rationalize the denominator of \frac{\sqrt{11}-3}{\sqrt{11}+3} by multiplying numerator and denominator by \sqrt{11}-3.

$$10+3\sqrt{11}+\frac{\left(\sqrt{11}-3\right)\left(\sqrt{11}-3\right)}{\left(\sqrt{11}+3\right)\left(\sqrt{11}-3\right)}$$

Consider \left(\sqrt{11}+3\right)\left(\sqrt{11}-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$10+3\sqrt{11}+\frac{\left(\sqrt{11}-3\right)\left(\sqrt{11}-3\right)}{\left(\sqrt{11}\right)^{2}-3^{2}}$$

Square \sqrt{11}. Square 3.

$$10+3\sqrt{11}+\frac{\left(\sqrt{11}-3\right)\left(\sqrt{11}-3\right)}{11-9}$$

Subtract 9 from 11 to get 2.

$$10+3\sqrt{11}+\frac{\left(\sqrt{11}-3\right)\left(\sqrt{11}-3\right)}{2}$$

Multiply \sqrt{11}-3 and \sqrt{11}-3 to get \left(\sqrt{11}-3\right)^{2}.

$$10+3\sqrt{11}+\frac{\left(\sqrt{11}-3\right)^{2}}{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{11}-3\right)^{2}.

$$10+3\sqrt{11}+\frac{\left(\sqrt{11}\right)^{2}-6\sqrt{11}+9}{2}$$

The square of \sqrt{11} is 11.

$$10+3\sqrt{11}+\frac{11-6\sqrt{11}+9}{2}$$

Add 11 and 9 to get 20.

$$10+3\sqrt{11}+\frac{20-6\sqrt{11}}{2}$$

Divide each term of 20-6\sqrt{11} by 2 to get 10-3\sqrt{11}.

$$10+3\sqrt{11}+10-3\sqrt{11}$$

Add 10 and 10 to get 20.

$$20+3\sqrt{11}-3\sqrt{11}$$

Combine 3\sqrt{11} and -3\sqrt{11} to get 0.

$$20$$

Factor

$$2^{2}\times 5$$