Solve frac{2sqrt{3}+8}{3+sqrt{3}}=

$$\frac { 2 \sqrt { 3 } + 8 } { 3 + \sqrt { 3 } } =$$

Evaluate

$$-\frac{\sqrt{3}}{3}+3\approx 2.422649731$$

Solution Steps

Rationalize the denominator of \frac{2\sqrt{3}+8}{3+\sqrt{3}} by multiplying numerator and denominator by 3-\sqrt{3}.

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

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

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

Square 3. Square \sqrt{3}.

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

Subtract 3 from 9 to get 6.

$$\frac{\left(2\sqrt{3}+8\right)\left(3-\sqrt{3}\right)}{6}$$

Apply the distributive property by multiplying each term of 2\sqrt{3}+8 by each term of 3-\sqrt{3}.

$$\frac{6\sqrt{3}-2\left(\sqrt{3}\right)^{2}+24-8\sqrt{3}}{6}$$

The square of \sqrt{3} is 3.

$$\frac{6\sqrt{3}-2\times 3+24-8\sqrt{3}}{6}$$

Multiply -2 and 3 to get -6.

$$\frac{6\sqrt{3}-6+24-8\sqrt{3}}{6}$$

Add -6 and 24 to get 18.

$$\frac{6\sqrt{3}+18-8\sqrt{3}}{6}$$

Combine 6\sqrt{3} and -8\sqrt{3} to get -2\sqrt{3}.

$$\frac{-2\sqrt{3}+18}{6}$$

Factor

$$\frac{9 - \sqrt{3}}{3} = 2.4226497308103743$$