Solve 7/-6-sqrt{3}

$$\frac { 7 } { - 6 - \sqrt { 3 } }$$

Evaluate

$$\frac{7\left(\sqrt{3}-6\right)}{33}\approx -0.905322556$$

Solution Steps

Rationalize the denominator of \frac{7}{-6-\sqrt{3}} by multiplying numerator and denominator by -6+\sqrt{3}.

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

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

$$\frac{7\left(-6+\sqrt{3}\right)}{\left(-6\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Square -6. Square \sqrt{3}.

$$\frac{7\left(-6+\sqrt{3}\right)}{36-3}$$

Subtract 3 from 36 to get 33.

$$\frac{7\left(-6+\sqrt{3}\right)}{33}$$

Use the distributive property to multiply 7 by -6+\sqrt{3}.

$$\frac{-42+7\sqrt{3}}{33}$$