How to solve 1/(7 + 3sqrt(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 1/(7 + 3sqrt(3)) | Equatix

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Question

$$\frac{1}{7+3\sqrt{3}}$$

Evaluate

$\frac{7-3\sqrt{3}}{22}\approx 0.081993072$

Solution Steps

Rationalize the denominator of $\frac{1}{7+3\sqrt{3}}$ by multiplying numerator and denominator by $7-3\sqrt{3}$.

$$\frac{7-3\sqrt{3}}{\left(7+3\sqrt{3}\right)\left(7-3\sqrt{3}\right)}$$

Consider $\left(7+3\sqrt{3}\right)\left(7-3\sqrt{3}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{7-3\sqrt{3}}{7^{2}-\left(3\sqrt{3}\right)^{2}}$$

Calculate $7$ to the power of $2$ and get $49$.

$$\frac{7-3\sqrt{3}}{49-\left(3\sqrt{3}\right)^{2}}$$

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

$$\frac{7-3\sqrt{3}}{49-3^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{7-3\sqrt{3}}{49-9\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{7-3\sqrt{3}}{49-9\times 3}$$

Multiply $9$ and $3$ to get $27$.

$$\frac{7-3\sqrt{3}}{49-27}$$

Subtract $27$ from $49$ to get $22$.

$$\frac{7-3\sqrt{3}}{22}$$

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