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

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

Evaluate

$$\frac{24\sqrt{3}-43}{11}\approx -0.130070965$$

Solution Steps

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

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

Consider \left(3\sqrt{3}+4\right)\left(3\sqrt{3}-4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

$$-\frac{9\left(\sqrt{3}\right)^{2}-24\sqrt{3}+16}{\left(3\sqrt{3}\right)^{2}-4^{2}}$$

The square of \sqrt{3} is 3.

$$-\frac{9\times 3-24\sqrt{3}+16}{\left(3\sqrt{3}\right)^{2}-4^{2}}$$

Multiply 9 and 3 to get 27.

$$-\frac{27-24\sqrt{3}+16}{\left(3\sqrt{3}\right)^{2}-4^{2}}$$

Add 27 and 16 to get 43.

$$-\frac{43-24\sqrt{3}}{\left(3\sqrt{3}\right)^{2}-4^{2}}$$

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

$$-\frac{43-24\sqrt{3}}{3^{2}\left(\sqrt{3}\right)^{2}-4^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$-\frac{43-24\sqrt{3}}{9\left(\sqrt{3}\right)^{2}-4^{2}}$$

The square of \sqrt{3} is 3.

$$-\frac{43-24\sqrt{3}}{9\times 3-4^{2}}$$

Multiply 9 and 3 to get 27.

$$-\frac{43-24\sqrt{3}}{27-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$-\frac{43-24\sqrt{3}}{27-16}$$

Subtract 16 from 27 to get 11.

$$-\frac{43-24\sqrt{3}}{11}$$