Solve frac{2sqrt{3}-12}{2sqrt{3}+6}

$$\frac { 2 \sqrt { 3 } - 12 } { 2 \sqrt { 3 } + 6 }$$

Evaluate

$$\frac{3\sqrt{3}-7}{2}\approx -0.901923789$$

Solution Steps

Rationalize the denominator of \frac{2\sqrt{3}-12}{2\sqrt{3}+6} by multiplying numerator and denominator by 2\sqrt{3}-6.

$$\frac{\left(2\sqrt{3}-12\right)\left(2\sqrt{3}-6\right)}{\left(2\sqrt{3}+6\right)\left(2\sqrt{3}-6\right)}$$

Consider \left(2\sqrt{3}+6\right)\left(2\sqrt{3}-6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2\sqrt{3}-12\right)\left(2\sqrt{3}-6\right)}{\left(2\sqrt{3}\right)^{2}-6^{2}}$$

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

$$\frac{\left(2\sqrt{3}-12\right)\left(2\sqrt{3}-6\right)}{2^{2}\left(\sqrt{3}\right)^{2}-6^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(2\sqrt{3}-12\right)\left(2\sqrt{3}-6\right)}{4\left(\sqrt{3}\right)^{2}-6^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(2\sqrt{3}-12\right)\left(2\sqrt{3}-6\right)}{4\times 3-6^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{\left(2\sqrt{3}-12\right)\left(2\sqrt{3}-6\right)}{12-6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{\left(2\sqrt{3}-12\right)\left(2\sqrt{3}-6\right)}{12-36}$$

Subtract 36 from 12 to get -24.

$$\frac{\left(2\sqrt{3}-12\right)\left(2\sqrt{3}-6\right)}{-24}$$

Apply the distributive property by multiplying each term of 2\sqrt{3}-12 by each term of 2\sqrt{3}-6.

$$\frac{4\left(\sqrt{3}\right)^{2}-12\sqrt{3}-24\sqrt{3}+72}{-24}$$

The square of \sqrt{3} is 3.

$$\frac{4\times 3-12\sqrt{3}-24\sqrt{3}+72}{-24}$$

Multiply 4 and 3 to get 12.

$$\frac{12-12\sqrt{3}-24\sqrt{3}+72}{-24}$$

Combine -12\sqrt{3} and -24\sqrt{3} to get -36\sqrt{3}.

$$\frac{12-36\sqrt{3}+72}{-24}$$

Add 12 and 72 to get 84.

$$\frac{84-36\sqrt{3}}{-24}$$

Factor

$$\frac{3 \sqrt{3} - 7}{2} = -0.901923788646684$$