How to solve 1/(1 + 3sqrt(2))

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/(1 + 3sqrt(2)) | Equatix

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Question

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

Evaluate

$\frac{3\sqrt{2}-1}{17}\approx 0.19074357$

Solution Steps

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

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

Consider $\left(1+3\sqrt{2}\right)\left(1-3\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

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

Calculate $1$ to the power of $2$ and get $1$.

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

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

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

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

$$\frac{1-3\sqrt{2}}{1-9\left(\sqrt{2}\right)^{2}}$$

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

$$\frac{1-3\sqrt{2}}{1-9\times 2}$$

Multiply $9$ and $2$ to get $18$.

$$\frac{1-3\sqrt{2}}{1-18}$$

Subtract $18$ from $1$ to get $-17$.

$$\frac{1-3\sqrt{2}}{-17}$$

Multiply both numerator and denominator by -1.

$$\frac{-1+3\sqrt{2}}{17}$$

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