Solve frac{(sqrt{5}+sqrt{3})(sqrt{5}-sqrt{3})}{2-sqrt{3}}

$$\frac{ ( \sqrt{ 5 } + \sqrt{ 3 } )( \sqrt{ 5 } - \sqrt{ 3 } ) }{ 2- \sqrt{ 3 } }$$

Evaluate

$$2\left(\sqrt{3}+2\right)\approx 7.464101615$$

Solution Steps

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}.

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

The square of \sqrt{5} is 5.

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

The square of \sqrt{3} is 3.

$$\frac{5-3}{2-\sqrt{3}}$$

Subtract 3 from 5 to get 2.

$$\frac{2}{2-\sqrt{3}}$$

Rationalize the denominator of \frac{2}{2-\sqrt{3}} by multiplying numerator and denominator by 2+\sqrt{3}.

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

Consider \left(2-\sqrt{3}\right)\left(2+\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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

$$\frac{2\left(2+\sqrt{3}\right)}{1}$$

Anything divided by one gives itself.

$$2\left(2+\sqrt{3}\right)$$

Use the distributive property to multiply 2 by 2+\sqrt{3}.

$$4+2\sqrt{3}$$