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.

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Question

$$\frac { 1 } { \sqrt { 9 } - \sqrt { 3 } }$$

Evaluate

$\frac{\sqrt{3}}{6}+\frac{1}{2}\approx 0.788675135$

Solution Steps

Calculate the square root of $9$ and get $3$.

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

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

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

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

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

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

$$\frac{3+\sqrt{3}}{9-3}$$

Subtract $3$ from $9$ to get $6$.

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

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