Solve 4sqrt{3}-6-1/4sqrt{3-6}-1

$$4 \sqrt { 3 } - 6 - \frac { 1 } { 4 \sqrt { 3 } - 6 } - 1$$

Evaluate

$$\frac{11\sqrt{3}}{3}-\frac{15}{2}\approx -1.149147039$$

Short Solution Steps

Rationalize the denominator of \frac{1}{4\sqrt{3}-6} by multiplying numerator and denominator by 4\sqrt{3}+6.

$$4\sqrt{3}-6-\frac{4\sqrt{3}+6}{\left(4\sqrt{3}-6\right)\left(4\sqrt{3}+6\right)}-1$$

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

$$4\sqrt{3}-6-\frac{4\sqrt{3}+6}{\left(4\sqrt{3}\right)^{2}-6^{2}}-1$$

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

$$4\sqrt{3}-6-\frac{4\sqrt{3}+6}{4^{2}\left(\sqrt{3}\right)^{2}-6^{2}}-1$$

Calculate 4 to the power of 2 and get 16.

$$4\sqrt{3}-6-\frac{4\sqrt{3}+6}{16\left(\sqrt{3}\right)^{2}-6^{2}}-1$$

The square of \sqrt{3} is 3.

$$4\sqrt{3}-6-\frac{4\sqrt{3}+6}{16\times 3-6^{2}}-1$$

Multiply 16 and 3 to get 48.

$$4\sqrt{3}-6-\frac{4\sqrt{3}+6}{48-6^{2}}-1$$

Calculate 6 to the power of 2 and get 36.

$$4\sqrt{3}-6-\frac{4\sqrt{3}+6}{48-36}-1$$

Subtract 36 from 48 to get 12.

$$4\sqrt{3}-6-\frac{4\sqrt{3}+6}{12}-1$$

Subtract 1 from -6 to get -7.

$$4\sqrt{3}-7-\frac{4\sqrt{3}+6}{12}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4\sqrt{3}-7 times \frac{12}{12}.

$$\frac{12\left(4\sqrt{3}-7\right)}{12}-\frac{4\sqrt{3}+6}{12}$$

Since \frac{12\left(4\sqrt{3}-7\right)}{12} and \frac{4\sqrt{3}+6}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{12\left(4\sqrt{3}-7\right)-\left(4\sqrt{3}+6\right)}{12}$$

Do the multiplications in 12\left(4\sqrt{3}-7\right)-\left(4\sqrt{3}+6\right).

$$\frac{48\sqrt{3}-84-4\sqrt{3}-6}{12}$$

Do the calculations in 48\sqrt{3}-84-4\sqrt{3}-6.

$$\frac{44\sqrt{3}-90}{12}$$

Factor

$$\frac{22 \sqrt{3} - 45}{6} = -1.1491470389141167$$