Solve 1/sqrt{3+12}

$$\frac { 1 } { \sqrt { 3 } + 12 }$$

Evaluate

$$-\frac{\sqrt{3}}{141}+\frac{4}{47}\approx 0.072822335$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{3}+12} by multiplying numerator and denominator by \sqrt{3}-12.

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

Consider \left(\sqrt{3}+12\right)\left(\sqrt{3}-12\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{3}-12}{\left(\sqrt{3}\right)^{2}-12^{2}}$$

Square \sqrt{3}. Square 12.

$$\frac{\sqrt{3}-12}{3-144}$$

Subtract 144 from 3 to get -141.

$$\frac{\sqrt{3}-12}{-141}$$

Multiply both numerator and denominator by -1.

$$\frac{-\sqrt{3}+12}{141}$$