Solve 6/sqrt{3-sqrt{2}}+6/sqrt{3+sqrt{2}}

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

Evaluate

$$12\sqrt{3}\approx 20.784609691$$

Solution Steps

Rationalize the denominator of \frac{6}{\sqrt{3}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}+\sqrt{2}.

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

Consider \left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

$$\frac{6\left(\sqrt{3}+\sqrt{2}\right)}{1}+\frac{6}{\sqrt{3}+\sqrt{2}}$$

Anything divided by one gives itself.

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

Rationalize the denominator of \frac{6}{\sqrt{3}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}-\sqrt{2}.

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

Consider \left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

$$6\left(\sqrt{3}+\sqrt{2}\right)+\frac{6\left(\sqrt{3}-\sqrt{2}\right)}{1}$$

Anything divided by one gives itself.

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

Use the distributive property to multiply 6 by \sqrt{3}+\sqrt{2}.

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

Use the distributive property to multiply 6 by \sqrt{3}-\sqrt{2}.

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

Combine 6\sqrt{3} and 6\sqrt{3} to get 12\sqrt{3}.

$$12\sqrt{3}+6\sqrt{2}-6\sqrt{2}$$

Combine 6\sqrt{2} and -6\sqrt{2} to get 0.

$$12\sqrt{3}$$