Solve 3/2sqrt{3+3}

$$\frac { 3 } { 2 \sqrt { 3 } + 3 }$$

Evaluate

$$2\sqrt{3}-3\approx 0.464101615$$

Solution Steps

Rationalize the denominator of \frac{3}{2\sqrt{3}+3} by multiplying numerator and denominator by 2\sqrt{3}-3.

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

Consider \left(2\sqrt{3}+3\right)\left(2\sqrt{3}-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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{3\left(2\sqrt{3}-3\right)}{12-3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{3\left(2\sqrt{3}-3\right)}{12-9}$$

Subtract 9 from 12 to get 3.

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

Cancel out 3 and 3.

$$2\sqrt{3}-3$$