Solve frac{8+3sqrt{7}}{8-3sqrt{7}}

$$\frac { 8 + 3 \sqrt { 7 } } { 8 - 3 \sqrt { 7 } }$$

Evaluate

$$48\sqrt{7}+127\approx 253.996062931$$

Solution Steps

Rationalize the denominator of \frac{8+3\sqrt{7}}{8-3\sqrt{7}} by multiplying numerator and denominator by 8+3\sqrt{7}.

$$\frac{\left(8+3\sqrt{7}\right)\left(8+3\sqrt{7}\right)}{\left(8-3\sqrt{7}\right)\left(8+3\sqrt{7}\right)}$$

Consider \left(8-3\sqrt{7}\right)\left(8+3\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(8+3\sqrt{7}\right)\left(8+3\sqrt{7}\right)}{8^{2}-\left(-3\sqrt{7}\right)^{2}}$$

Multiply 8+3\sqrt{7} and 8+3\sqrt{7} to get \left(8+3\sqrt{7}\right)^{2}.

$$\frac{\left(8+3\sqrt{7}\right)^{2}}{8^{2}-\left(-3\sqrt{7}\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8+3\sqrt{7}\right)^{2}.

$$\frac{64+48\sqrt{7}+9\left(\sqrt{7}\right)^{2}}{8^{2}-\left(-3\sqrt{7}\right)^{2}}$$

The square of \sqrt{7} is 7.

$$\frac{64+48\sqrt{7}+9\times 7}{8^{2}-\left(-3\sqrt{7}\right)^{2}}$$

Multiply 9 and 7 to get 63.

$$\frac{64+48\sqrt{7}+63}{8^{2}-\left(-3\sqrt{7}\right)^{2}}$$

Add 64 and 63 to get 127.

$$\frac{127+48\sqrt{7}}{8^{2}-\left(-3\sqrt{7}\right)^{2}}$$

Calculate 8 to the power of 2 and get 64.

$$\frac{127+48\sqrt{7}}{64-\left(-3\sqrt{7}\right)^{2}}$$

Expand \left(-3\sqrt{7}\right)^{2}.

$$\frac{127+48\sqrt{7}}{64-\left(-3\right)^{2}\left(\sqrt{7}\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$\frac{127+48\sqrt{7}}{64-9\left(\sqrt{7}\right)^{2}}$$

The square of \sqrt{7} is 7.

$$\frac{127+48\sqrt{7}}{64-9\times 7}$$

Multiply 9 and 7 to get 63.

$$\frac{127+48\sqrt{7}}{64-63}$$

Subtract 63 from 64 to get 1.

$$\frac{127+48\sqrt{7}}{1}$$

Anything divided by one gives itself.

$$127+48\sqrt{7}$$