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

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

Evaluate

$$\frac{19-5\sqrt{3}}{22}\approx 0.469988453$$

Solution Steps

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

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

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

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

Calculate 7 to the power of 2 and get 49.

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

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

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

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(4+\sqrt{3}\right)\left(7-3\sqrt{3}\right)}{49-9\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(4+\sqrt{3}\right)\left(7-3\sqrt{3}\right)}{49-9\times 3}$$

Multiply 9 and 3 to get 27.

$$\frac{\left(4+\sqrt{3}\right)\left(7-3\sqrt{3}\right)}{49-27}$$

Subtract 27 from 49 to get 22.

$$\frac{\left(4+\sqrt{3}\right)\left(7-3\sqrt{3}\right)}{22}$$

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

$$\frac{28-12\sqrt{3}+7\sqrt{3}-3\left(\sqrt{3}\right)^{2}}{22}$$

Combine -12\sqrt{3} and 7\sqrt{3} to get -5\sqrt{3}.

$$\frac{28-5\sqrt{3}-3\left(\sqrt{3}\right)^{2}}{22}$$

The square of \sqrt{3} is 3.

$$\frac{28-5\sqrt{3}-3\times 3}{22}$$

Multiply -3 and 3 to get -9.

$$\frac{28-5\sqrt{3}-9}{22}$$

Subtract 9 from 28 to get 19.

$$\frac{19-5\sqrt{3}}{22}$$