Solve 1/-2sqrt{3-3}

$$\frac { 1 } { - 2 \sqrt { 3 } - 3 }$$

Evaluate

$$-\frac{2\sqrt{3}}{3}+1\approx -0.154700538$$

Solution Steps

Rationalize the denominator of \frac{1}{-2\sqrt{3}-3} by multiplying numerator and denominator by -2\sqrt{3}+3.

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

Consider \left(-2\sqrt{3}-3\right)\left(-2\sqrt{3}+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}.

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

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

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

Calculate -2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{-2\sqrt{3}+3}{12-9}$$

Subtract 9 from 12 to get 3.

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