Solve frac{1+4sqrt{3}}{1-4sqrt{6}}

$$\frac { 1 + 4 \sqrt { 3 } } { 1 - 4 \sqrt { 6 } }$$

Evaluate

$$\frac{-4\sqrt{3}-4\sqrt{6}-48\sqrt{2}-1}{95}\approx -0.901141192$$

Solution Steps

Rationalize the denominator of \frac{1+4\sqrt{3}}{1-4\sqrt{6}} by multiplying numerator and denominator by 1+4\sqrt{6}.

$$\frac{\left(1+4\sqrt{3}\right)\left(1+4\sqrt{6}\right)}{\left(1-4\sqrt{6}\right)\left(1+4\sqrt{6}\right)}$$

Consider \left(1-4\sqrt{6}\right)\left(1+4\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(1+4\sqrt{3}\right)\left(1+4\sqrt{6}\right)}{1^{2}-\left(-4\sqrt{6}\right)^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{\left(1+4\sqrt{3}\right)\left(1+4\sqrt{6}\right)}{1-\left(-4\sqrt{6}\right)^{2}}$$

Expand \left(-4\sqrt{6}\right)^{2}.

$$\frac{\left(1+4\sqrt{3}\right)\left(1+4\sqrt{6}\right)}{1-\left(-4\right)^{2}\left(\sqrt{6}\right)^{2}}$$

Calculate -4 to the power of 2 and get 16.

$$\frac{\left(1+4\sqrt{3}\right)\left(1+4\sqrt{6}\right)}{1-16\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{\left(1+4\sqrt{3}\right)\left(1+4\sqrt{6}\right)}{1-16\times 6}$$

Multiply 16 and 6 to get 96.

$$\frac{\left(1+4\sqrt{3}\right)\left(1+4\sqrt{6}\right)}{1-96}$$

Subtract 96 from 1 to get -95.

$$\frac{\left(1+4\sqrt{3}\right)\left(1+4\sqrt{6}\right)}{-95}$$

Apply the distributive property by multiplying each term of 1+4\sqrt{3} by each term of 1+4\sqrt{6}.

$$\frac{1+4\sqrt{6}+4\sqrt{3}+16\sqrt{3}\sqrt{6}}{-95}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{1+4\sqrt{6}+4\sqrt{3}+16\sqrt{3}\sqrt{3}\sqrt{2}}{-95}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{1+4\sqrt{6}+4\sqrt{3}+16\times 3\sqrt{2}}{-95}$$

Multiply 16 and 3 to get 48.

$$\frac{1+4\sqrt{6}+4\sqrt{3}+48\sqrt{2}}{-95}$$

Multiply both numerator and denominator by -1.

$$\frac{-1-4\sqrt{6}-4\sqrt{3}-48\sqrt{2}}{95}$$