Solve 3/4sqrt{3+3sqrt{5}}

$$\frac { 3 } { 4 \sqrt { 3 } + 3 \sqrt { 5 } }$$

Evaluate

$$4\sqrt{3}-3\sqrt{5}\approx 0.219999298$$

Solution Steps

Rationalize the denominator of \frac{3}{4\sqrt{3}+3\sqrt{5}} by multiplying numerator and denominator by 4\sqrt{3}-3\sqrt{5}.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{\left(4\sqrt{3}+3\sqrt{5}\right)\left(4\sqrt{3}-3\sqrt{5}\right)}$$

Consider \left(4\sqrt{3}+3\sqrt{5}\right)\left(4\sqrt{3}-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}.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{\left(4\sqrt{3}\right)^{2}-\left(3\sqrt{5}\right)^{2}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{4^{2}\left(\sqrt{3}\right)^{2}-\left(3\sqrt{5}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{16\left(\sqrt{3}\right)^{2}-\left(3\sqrt{5}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{16\times 3-\left(3\sqrt{5}\right)^{2}}$$

Multiply 16 and 3 to get 48.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{48-\left(3\sqrt{5}\right)^{2}}$$

Expand \left(3\sqrt{5}\right)^{2}.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{48-3^{2}\left(\sqrt{5}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{48-9\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{48-9\times 5}$$

Multiply 9 and 5 to get 45.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{48-45}$$

Subtract 45 from 48 to get 3.

$$\frac{3\left(4\sqrt{3}-3\sqrt{5}\right)}{3}$$

Cancel out 3 and 3.

$$4\sqrt{3}-3\sqrt{5}$$