Solve 40/1-sqrt{3}

$$\frac { 40 } { 1 - \sqrt { 3 } }$$

Evaluate

$$-20\sqrt{3}-20\approx -54.641016151$$

Solution Steps

Rationalize the denominator of \frac{40}{1-\sqrt{3}} by multiplying numerator and denominator by 1+\sqrt{3}.

$$\frac{40\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}$$

Consider \left(1-\sqrt{3}\right)\left(1+\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}.

$$\frac{40\left(1+\sqrt{3}\right)}{1^{2}-\left(\sqrt{3}\right)^{2}}$$

Square 1. Square \sqrt{3}.

$$\frac{40\left(1+\sqrt{3}\right)}{1-3}$$

Subtract 3 from 1 to get -2.

$$\frac{40\left(1+\sqrt{3}\right)}{-2}$$

Divide 40\left(1+\sqrt{3}\right) by -2 to get -20\left(1+\sqrt{3}\right).

$$-20\left(1+\sqrt{3}\right)$$

Use the distributive property to multiply -20 by 1+\sqrt{3}.

$$-20-20\sqrt{3}$$