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

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

Evaluate

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

Short Solution Steps

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

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

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

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

Multiply x\sqrt{3}+\sqrt{2} and x\sqrt{3}+\sqrt{2} to get \left(x\sqrt{3}+\sqrt{2}\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x\sqrt{3}+\sqrt{2}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{2} is 2.

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

Expand \left(x\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

The square of \sqrt{2} is 2.

$$\frac{x^{2}\times 3+2x\sqrt{6}+2}{x^{2}\times 3-2}+\frac{x\sqrt{3}-\sqrt{2}}{x\sqrt{3}+\sqrt{2}}$$

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

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

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

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

Multiply x\sqrt{3}-\sqrt{2} and x\sqrt{3}-\sqrt{2} to get \left(x\sqrt{3}-\sqrt{2}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x\sqrt{3}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{2} is 2.

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

Expand \left(x\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

The square of \sqrt{2} is 2.

$$\frac{x^{2}\times 3+2x\sqrt{6}+2}{x^{2}\times 3-2}+\frac{x^{2}\times 3-2x\sqrt{6}+2}{x^{2}\times 3-2}$$

Since \frac{x^{2}\times 3+2x\sqrt{6}+2}{x^{2}\times 3-2} and \frac{x^{2}\times 3-2x\sqrt{6}+2}{x^{2}\times 3-2} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}\times 3+2x\sqrt{6}+2+x^{2}\times 3-2x\sqrt{6}+2}{x^{2}\times 3-2}$$

Combine like terms in x^{2}\times 3+2x\sqrt{6}+2+x^{2}\times 3-2x\sqrt{6}+2.

$$\frac{6x^{2}+4}{x^{2}\times 3-2}$$

Factor

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