Solve frac{5+sqrt{3}}{7-4sqrt{3}}-frac{5-sqrt{3}}{7+4sqrt{3}}

$$\frac { 5 + \sqrt { 3 } } { 7 - 4 \sqrt { 3 } } - \frac { 5 - \sqrt { 3 } } { 7 + 4 \sqrt { 3 } }$$

Evaluate

$$54\sqrt{3}\approx 93.530743609$$

Solution Steps

Rationalize the denominator of \frac{5+\sqrt{3}}{7-4\sqrt{3}} by multiplying numerator and denominator by 7+4\sqrt{3}.

$$\frac{\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)}{\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)}-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

Consider \left(7-4\sqrt{3}\right)\left(7+4\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}.

$$\frac{\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)}{7^{2}-\left(-4\sqrt{3}\right)^{2}}-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)}{49-\left(-4\sqrt{3}\right)^{2}}-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

Expand \left(-4\sqrt{3}\right)^{2}.

$$\frac{\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)}{49-\left(-4\right)^{2}\left(\sqrt{3}\right)^{2}}-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

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

$$\frac{\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)}{49-16\times 3}-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

Multiply 16 and 3 to get 48.

$$\frac{\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)}{49-48}-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

Subtract 48 from 49 to get 1.

$$\frac{\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)}{1}-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

Anything divided by one gives itself.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{5-\sqrt{3}}{7+4\sqrt{3}}$$

Rationalize the denominator of \frac{5-\sqrt{3}}{7+4\sqrt{3}} by multiplying numerator and denominator by 7-4\sqrt{3}.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\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}.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{7^{2}-\left(4\sqrt{3}\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-\left(4\sqrt{3}\right)^{2}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-4^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\times 3}$$

Multiply 16 and 3 to get 48.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-48}$$

Subtract 48 from 49 to get 1.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\frac{\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{1}$$

Anything divided by one gives itself.

$$\left(5+\sqrt{3}\right)\left(7+4\sqrt{3}\right)-\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

Apply the distributive property by multiplying each term of 5+\sqrt{3} by each term of 7+4\sqrt{3}.

$$35+20\sqrt{3}+7\sqrt{3}+4\left(\sqrt{3}\right)^{2}-\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

Combine 20\sqrt{3} and 7\sqrt{3} to get 27\sqrt{3}.

$$35+27\sqrt{3}+4\left(\sqrt{3}\right)^{2}-\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$35+27\sqrt{3}+4\times 3-\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

Multiply 4 and 3 to get 12.

$$35+27\sqrt{3}+12-\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

Add 35 and 12 to get 47.

$$47+27\sqrt{3}-\left(5-\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

Apply the distributive property by multiplying each term of 5-\sqrt{3} by each term of 7-4\sqrt{3}.

$$47+27\sqrt{3}-\left(35-20\sqrt{3}-7\sqrt{3}+4\left(\sqrt{3}\right)^{2}\right)$$

Combine -20\sqrt{3} and -7\sqrt{3} to get -27\sqrt{3}.

$$47+27\sqrt{3}-\left(35-27\sqrt{3}+4\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$47+27\sqrt{3}-\left(35-27\sqrt{3}+4\times 3\right)$$

Multiply 4 and 3 to get 12.

$$47+27\sqrt{3}-\left(35-27\sqrt{3}+12\right)$$

Add 35 and 12 to get 47.

$$47+27\sqrt{3}-\left(47-27\sqrt{3}\right)$$

To find the opposite of 47-27\sqrt{3}, find the opposite of each term.

$$47+27\sqrt{3}-47-\left(-27\sqrt{3}\right)$$

The opposite of -27\sqrt{3} is 27\sqrt{3}.

$$47+27\sqrt{3}-47+27\sqrt{3}$$

Subtract 47 from 47 to get 0.

$$27\sqrt{3}+27\sqrt{3}$$

Combine 27\sqrt{3} and 27\sqrt{3} to get 54\sqrt{3}.

$$54\sqrt{3}$$