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

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

Evaluate

$$\sqrt{15}+4\sqrt{3}-2\sqrt{5}-6\approx 0.329050621$$

Solution Steps

Rationalize the denominator of \frac{2\sqrt{3}-\sqrt{5}}{2+\sqrt{3}} by multiplying numerator and denominator by 2-\sqrt{3}.

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

$$\frac{\left(2\sqrt{3}-\sqrt{5}\right)\left(2-\sqrt{3}\right)}{1}$$

Anything divided by one gives itself.

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

Apply the distributive property by multiplying each term of 2\sqrt{3}-\sqrt{5} by each term of 2-\sqrt{3}.

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

The square of \sqrt{3} is 3.

$$4\sqrt{3}-2\times 3-2\sqrt{5}+\sqrt{3}\sqrt{5}$$

Multiply -2 and 3 to get -6.

$$4\sqrt{3}-6-2\sqrt{5}+\sqrt{3}\sqrt{5}$$

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

$$4\sqrt{3}-6-2\sqrt{5}+\sqrt{15}$$