Solve 8/2sqrt{3+2}

$$\frac { 8 } { 2 \sqrt { 3 } + 2 }$$

Evaluate

$$2\left(\sqrt{3}-1\right)\approx 1.464101615$$

Solution Steps

Rationalize the denominator of \frac{8}{2\sqrt{3}+2} by multiplying numerator and denominator by 2\sqrt{3}-2.

$$\frac{8\left(2\sqrt{3}-2\right)}{\left(2\sqrt{3}+2\right)\left(2\sqrt{3}-2\right)}$$

Consider \left(2\sqrt{3}+2\right)\left(2\sqrt{3}-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{8\left(2\sqrt{3}-2\right)}{\left(2\sqrt{3}\right)^{2}-2^{2}}$$

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

$$\frac{8\left(2\sqrt{3}-2\right)}{2^{2}\left(\sqrt{3}\right)^{2}-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{8\left(2\sqrt{3}-2\right)}{12-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{8\left(2\sqrt{3}-2\right)}{12-4}$$

Subtract 4 from 12 to get 8.

$$\frac{8\left(2\sqrt{3}-2\right)}{8}$$

Cancel out 8 and 8.

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