How to solve 3 12+√3

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 3 12+√3 | Equatix

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Question

$$\frac { 1 } { \sqrt { 2 } + \sqrt { 3 } }$$

Evaluate

$\sqrt{3}-\sqrt{2}\approx 0.317837245$

Solution Steps

Rationalize the denominator of $\frac{1}{\sqrt{2}+\sqrt{3}}$ by multiplying numerator and denominator by $\sqrt{2}-\sqrt{3}$.

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

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

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

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

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

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

$$\frac{\sqrt{2}-\sqrt{3}}{-1}$$

Anything divided by -1 gives its opposite. To find the opposite of $\sqrt{2}-\sqrt{3}$, find the opposite of each term.

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

The opposite of $-\sqrt{3}$ is $\sqrt{3}$.

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

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