Solve 5/7sqrt{3+sqrt{5}}

$$\frac { 5 } { 7 \sqrt { 3 } + \sqrt { 5 } }$$

Evaluate

$$\frac{35\sqrt{3}-5\sqrt{5}}{142}\approx 0.348179144$$

Solution Steps

Rationalize the denominator of \frac{5}{7\sqrt{3}+\sqrt{5}} by multiplying numerator and denominator by 7\sqrt{3}-\sqrt{5}.

$$\frac{5\left(7\sqrt{3}-\sqrt{5}\right)}{\left(7\sqrt{3}+\sqrt{5}\right)\left(7\sqrt{3}-\sqrt{5}\right)}$$

Consider \left(7\sqrt{3}+\sqrt{5}\right)\left(7\sqrt{3}-\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}.

$$\frac{5\left(7\sqrt{3}-\sqrt{5}\right)}{\left(7\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}}$$

Expand \left(7\sqrt{3}\right)^{2}.

$$\frac{5\left(7\sqrt{3}-\sqrt{5}\right)}{7^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{5\left(7\sqrt{3}-\sqrt{5}\right)}{49\left(\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{5\left(7\sqrt{3}-\sqrt{5}\right)}{49\times 3-\left(\sqrt{5}\right)^{2}}$$

Multiply 49 and 3 to get 147.

$$\frac{5\left(7\sqrt{3}-\sqrt{5}\right)}{147-\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{5\left(7\sqrt{3}-\sqrt{5}\right)}{147-5}$$

Subtract 5 from 147 to get 142.

$$\frac{5\left(7\sqrt{3}-\sqrt{5}\right)}{142}$$

Use the distributive property to multiply 5 by 7\sqrt{3}-\sqrt{5}.

$$\frac{35\sqrt{3}-5\sqrt{5}}{142}$$