Solve frac{sqrt{3}+1}{sqrt{3}-4}

$$\frac { \sqrt { 3 } + 1 } { \sqrt { 3 } - 4 }$$

Evaluate

$$\frac{-5\sqrt{3}-7}{13}\approx -1.204634926$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{3}+1}{\sqrt{3}-4} by multiplying numerator and denominator by \sqrt{3}+4.

$$\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}+4\right)}{\left(\sqrt{3}-4\right)\left(\sqrt{3}+4\right)}$$

Consider \left(\sqrt{3}-4\right)\left(\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}.

$$\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}+4\right)}{\left(\sqrt{3}\right)^{2}-4^{2}}$$

Square \sqrt{3}. Square 4.

$$\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}+4\right)}{3-16}$$

Subtract 16 from 3 to get -13.

$$\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}+4\right)}{-13}$$

Apply the distributive property by multiplying each term of \sqrt{3}+1 by each term of \sqrt{3}+4.

$$\frac{\left(\sqrt{3}\right)^{2}+4\sqrt{3}+\sqrt{3}+4}{-13}$$

The square of \sqrt{3} is 3.

$$\frac{3+4\sqrt{3}+\sqrt{3}+4}{-13}$$

Combine 4\sqrt{3} and \sqrt{3} to get 5\sqrt{3}.

$$\frac{3+5\sqrt{3}+4}{-13}$$

Add 3 and 4 to get 7.

$$\frac{7+5\sqrt{3}}{-13}$$

Multiply both numerator and denominator by -1.

$$\frac{-7-5\sqrt{3}}{13}$$