Solve frac{sqrt{7}+sqrt{5}}{sqrt{7}-sqrt{5}}-sqrt{35}

$$\frac{ \sqrt{ 7 } + \sqrt{ 5 } }{ \sqrt{ 7 } - \sqrt{ 5 } } - \sqrt{ 35 }$$

Evaluate

$$6$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} by multiplying numerator and denominator by \sqrt{7}+\sqrt{5}.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}-\sqrt{35}$$

Consider \left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{5}\right)^{2}}-\sqrt{35}$$

Square \sqrt{7}. Square \sqrt{5}.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{7-5}-\sqrt{35}$$

Subtract 5 from 7 to get 2.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}{2}-\sqrt{35}$$

Multiply \sqrt{7}+\sqrt{5} and \sqrt{7}+\sqrt{5} to get \left(\sqrt{7}+\sqrt{5}\right)^{2}.

$$\frac{\left(\sqrt{7}+\sqrt{5}\right)^{2}}{2}-\sqrt{35}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+\sqrt{5}\right)^{2}.

$$\frac{\left(\sqrt{7}\right)^{2}+2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}}{2}-\sqrt{35}$$

The square of \sqrt{7} is 7.

$$\frac{7+2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}}{2}-\sqrt{35}$$

To multiply \sqrt{7} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{7+2\sqrt{35}+\left(\sqrt{5}\right)^{2}}{2}-\sqrt{35}$$

The square of \sqrt{5} is 5.

$$\frac{7+2\sqrt{35}+5}{2}-\sqrt{35}$$

Add 7 and 5 to get 12.

$$\frac{12+2\sqrt{35}}{2}-\sqrt{35}$$

Divide each term of 12+2\sqrt{35} by 2 to get 6+\sqrt{35}.

$$6+\sqrt{35}-\sqrt{35}$$

Combine \sqrt{35} and -\sqrt{35} to get 0.

$$6$$

Factor

$$2\times 3$$