Solve frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}-frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}=2sqrt{15}

$$\frac { \sqrt { 5 } + \sqrt { 3 } } { \sqrt { 5 } - \sqrt { 3 } } - \frac { \sqrt { 5 } - \sqrt { 3 } } { \sqrt { 5 } + \sqrt { 3 } } = 2 \sqrt { 15 }$$

Verify

Solution Steps

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

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

Consider \left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\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}.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

Square \sqrt{5}. Square \sqrt{3}.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

Subtract 3 from 5 to get 2.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

Multiply \sqrt{5}+\sqrt{3} and \sqrt{5}+\sqrt{3} to get \left(\sqrt{5}+\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

The square of \sqrt{5} is 5.

$$\frac{5+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{5+2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

The square of \sqrt{3} is 3.

$$\frac{5+2\sqrt{15}+3}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

Add 5 and 3 to get 8.

$$\frac{8+2\sqrt{15}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

Divide each term of 8+2\sqrt{15} by 2 to get 4+\sqrt{15}.

$$4+\sqrt{15}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}$$

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

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=2\sqrt{15}$$

Consider \left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\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}.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}=2\sqrt{15}$$

Square \sqrt{5}. Square \sqrt{3}.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{5-3}=2\sqrt{15}$$

Subtract 3 from 5 to get 2.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}=2\sqrt{15}$$

Multiply \sqrt{5}-\sqrt{3} and \sqrt{5}-\sqrt{3} to get \left(\sqrt{5}-\sqrt{3}\right)^{2}.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{2}=2\sqrt{15}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-\sqrt{3}\right)^{2}.

$$4+\sqrt{15}-\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}=2\sqrt{15}$$

The square of \sqrt{5} is 5.

$$4+\sqrt{15}-\frac{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}=2\sqrt{15}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$4+\sqrt{15}-\frac{5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}=2\sqrt{15}$$

The square of \sqrt{3} is 3.

$$4+\sqrt{15}-\frac{5-2\sqrt{15}+3}{2}=2\sqrt{15}$$

Add 5 and 3 to get 8.

$$4+\sqrt{15}-\frac{8-2\sqrt{15}}{2}=2\sqrt{15}$$

Divide each term of 8-2\sqrt{15} by 2 to get 4-\sqrt{15}.

$$4+\sqrt{15}-\left(4-\sqrt{15}\right)=2\sqrt{15}$$

To find the opposite of 4-\sqrt{15}, find the opposite of each term.

$$4+\sqrt{15}-4-\left(-\sqrt{15}\right)=2\sqrt{15}$$

The opposite of -\sqrt{15} is \sqrt{15}.

$$4+\sqrt{15}-4+\sqrt{15}=2\sqrt{15}$$

Subtract 4 from 4 to get 0.

$$\sqrt{15}+\sqrt{15}=2\sqrt{15}$$

Combine \sqrt{15} and \sqrt{15} to get 2\sqrt{15}.

$$2\sqrt{15}=2\sqrt{15}$$

Subtract 2\sqrt{15} from both sides.

$$2\sqrt{15}-2\sqrt{15}=0$$

Combine 2\sqrt{15} and -2\sqrt{15} to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$