Solve (sqrt(3) + sqrt(2))/(sqrt(3) - sqrt(2)) + (sqrt(3) - sqrt(2))/(sqrt(3) + sqrt(2)) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Evaluate

$10$

Solution Steps

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

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

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

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

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

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

Subtract $2$ from $3$ to get $1$.

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

Anything divided by one gives itself.

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

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

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

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

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

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

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

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

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

Subtract $2$ from $3$ to get $1$.

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

Anything divided by one gives itself.

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

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

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

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

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

The square of $\sqrt{3}$ is $3$.

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

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

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

The square of $\sqrt{2}$ is $2$.

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

Add $3$ and $2$ to get $5$.

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

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

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

The square of $\sqrt{3}$ is $3$.

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

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

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

The square of $\sqrt{2}$ is $2$.

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

Add $3$ and $2$ to get $5$.

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

Add $5$ and $5$ to get $10$.

$$10+2\sqrt{6}-2\sqrt{6}$$

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

$$10$$

Factor

$2\times 5$