Solve 4/sqrt{11+sqrt{7}}-8/sqrt{11+sqrt{3}}+4/sqrt{7-sqrt{3}}

$$\frac { 4 } { \sqrt { 11 } + \sqrt { 7 } } - \frac { 8 } { \sqrt { 11 } + \sqrt { 3 } } + \frac { 4 } { \sqrt { 7 } - \sqrt { 3 } }$$

Evaluate

$$2\sqrt{3}\approx 3.464101615$$

Solution Steps

Rationalize the denominator of \frac{4}{\sqrt{11}+\sqrt{7}} by multiplying numerator and denominator by \sqrt{11}-\sqrt{7}.

$$\frac{4\left(\sqrt{11}-\sqrt{7}\right)}{\left(\sqrt{11}+\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right)}-\frac{8}{\sqrt{11}+\sqrt{3}}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Consider \left(\sqrt{11}+\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\left(\sqrt{11}-\sqrt{7}\right)}{\left(\sqrt{11}\right)^{2}-\left(\sqrt{7}\right)^{2}}-\frac{8}{\sqrt{11}+\sqrt{3}}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Square \sqrt{11}. Square \sqrt{7}.

$$\frac{4\left(\sqrt{11}-\sqrt{7}\right)}{11-7}-\frac{8}{\sqrt{11}+\sqrt{3}}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Subtract 7 from 11 to get 4.

$$\frac{4\left(\sqrt{11}-\sqrt{7}\right)}{4}-\frac{8}{\sqrt{11}+\sqrt{3}}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Cancel out 4 and 4.

$$\sqrt{11}-\sqrt{7}-\frac{8}{\sqrt{11}+\sqrt{3}}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Rationalize the denominator of \frac{8}{\sqrt{11}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{11}-\sqrt{3}.

$$\sqrt{11}-\sqrt{7}-\frac{8\left(\sqrt{11}-\sqrt{3}\right)}{\left(\sqrt{11}+\sqrt{3}\right)\left(\sqrt{11}-\sqrt{3}\right)}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Consider \left(\sqrt{11}+\sqrt{3}\right)\left(\sqrt{11}-\sqrt{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}.

$$\sqrt{11}-\sqrt{7}-\frac{8\left(\sqrt{11}-\sqrt{3}\right)}{\left(\sqrt{11}\right)^{2}-\left(\sqrt{3}\right)^{2}}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Square \sqrt{11}. Square \sqrt{3}.

$$\sqrt{11}-\sqrt{7}-\frac{8\left(\sqrt{11}-\sqrt{3}\right)}{11-3}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Subtract 3 from 11 to get 8.

$$\sqrt{11}-\sqrt{7}-\frac{8\left(\sqrt{11}-\sqrt{3}\right)}{8}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Cancel out 8 and 8.

$$\sqrt{11}-\sqrt{7}-\left(\sqrt{11}-\sqrt{3}\right)+\frac{4}{\sqrt{7}-\sqrt{3}}$$

To find the opposite of \sqrt{11}-\sqrt{3}, find the opposite of each term.

$$\sqrt{11}-\sqrt{7}-\sqrt{11}-\left(-\sqrt{3}\right)+\frac{4}{\sqrt{7}-\sqrt{3}}$$

The opposite of -\sqrt{3} is \sqrt{3}.

$$\sqrt{11}-\sqrt{7}-\sqrt{11}+\sqrt{3}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

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

$$-\sqrt{7}+\sqrt{3}+\frac{4}{\sqrt{7}-\sqrt{3}}$$

Rationalize the denominator of \frac{4}{\sqrt{7}-\sqrt{3}} by multiplying numerator and denominator by \sqrt{7}+\sqrt{3}.

$$-\sqrt{7}+\sqrt{3}+\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}$$

Consider \left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{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}.

$$-\sqrt{7}+\sqrt{3}+\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Square \sqrt{7}. Square \sqrt{3}.

$$-\sqrt{7}+\sqrt{3}+\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{7-3}$$

Subtract 3 from 7 to get 4.

$$-\sqrt{7}+\sqrt{3}+\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}$$

Cancel out 4 and 4.

$$-\sqrt{7}+\sqrt{3}+\sqrt{7}+\sqrt{3}$$

Combine -\sqrt{7} and \sqrt{7} to get 0.

$$\sqrt{3}+\sqrt{3}$$

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

$$2\sqrt{3}$$