Solve frac{6-4sqrt{3}}{6+4sqrt{3}}

$$\frac { 6 - 4 \sqrt { 3 } } { 6 + 4 \sqrt { 3 } }$$

Evaluate

$$4\sqrt{3}-7\approx -0.07179677$$

Solution Steps

Rationalize the denominator of \frac{6-4\sqrt{3}}{6+4\sqrt{3}} by multiplying numerator and denominator by 6-4\sqrt{3}.

$$\frac{\left(6-4\sqrt{3}\right)\left(6-4\sqrt{3}\right)}{\left(6+4\sqrt{3}\right)\left(6-4\sqrt{3}\right)}$$

Consider \left(6+4\sqrt{3}\right)\left(6-4\sqrt{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}.

$$\frac{\left(6-4\sqrt{3}\right)\left(6-4\sqrt{3}\right)}{6^{2}-\left(4\sqrt{3}\right)^{2}}$$

Multiply 6-4\sqrt{3} and 6-4\sqrt{3} to get \left(6-4\sqrt{3}\right)^{2}.

$$\frac{\left(6-4\sqrt{3}\right)^{2}}{6^{2}-\left(4\sqrt{3}\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-4\sqrt{3}\right)^{2}.

$$\frac{36-48\sqrt{3}+16\left(\sqrt{3}\right)^{2}}{6^{2}-\left(4\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{36-48\sqrt{3}+16\times 3}{6^{2}-\left(4\sqrt{3}\right)^{2}}$$

Multiply 16 and 3 to get 48.

$$\frac{36-48\sqrt{3}+48}{6^{2}-\left(4\sqrt{3}\right)^{2}}$$

Add 36 and 48 to get 84.

$$\frac{84-48\sqrt{3}}{6^{2}-\left(4\sqrt{3}\right)^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{84-48\sqrt{3}}{36-\left(4\sqrt{3}\right)^{2}}$$

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

$$\frac{84-48\sqrt{3}}{36-4^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{84-48\sqrt{3}}{36-16\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{84-48\sqrt{3}}{36-16\times 3}$$

Multiply 16 and 3 to get 48.

$$\frac{84-48\sqrt{3}}{36-48}$$

Subtract 48 from 36 to get -12.

$$\frac{84-48\sqrt{3}}{-12}$$

Divide each term of 84-48\sqrt{3} by -12 to get -7+4\sqrt{3}.

$$-7+4\sqrt{3}$$

Factor

$$-\left(7-4\sqrt{3}\right)$$