Solve 9/sqrt{3}+6/sqrt{5+sqrt{3}}=

$$\frac { 9 } { \sqrt { 3 } } + \frac { 6 } { \sqrt { 5 } + \sqrt { 3 } } =$$

Evaluate

$$3\sqrt{5}\approx 6.708203932$$

Solution Steps

Rationalize the denominator of \frac{9}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{9\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+\frac{6}{\sqrt{5}+\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$\frac{9\sqrt{3}}{3}+\frac{6}{\sqrt{5}+\sqrt{3}}$$

Divide 9\sqrt{3} by 3 to get 3\sqrt{3}.

$$3\sqrt{3}+\frac{6}{\sqrt{5}+\sqrt{3}}$$

Rationalize the denominator of \frac{6}{\sqrt{5}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{5}-\sqrt{3}.

$$3\sqrt{3}+\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}$$

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}.

$$3\sqrt{3}+\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

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

$$3\sqrt{3}+\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{5-3}$$

Subtract 3 from 5 to get 2.

$$3\sqrt{3}+\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{2}$$

Divide 6\left(\sqrt{5}-\sqrt{3}\right) by 2 to get 3\left(\sqrt{5}-\sqrt{3}\right).

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

Use the distributive property to multiply 3 by \sqrt{5}-\sqrt{3}.

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

Combine 3\sqrt{3} and -3\sqrt{3} to get 0.

$$3\sqrt{5}$$