How to solve 1 2sqrt(5) - 1

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 2sqrt(5) - 1 | Equatix

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Question

$$\frac { 3 \sqrt { 2 } + 1 } { 2 \sqrt { 5 } - 1 }$$

Evaluate

$\frac{2\sqrt{5}+3\sqrt{2}+6\sqrt{10}+1}{19}\approx 1.509918032$

Solution Steps

Rationalize the denominator of $\frac{3\sqrt{2}+1}{2\sqrt{5}-1}$ by multiplying numerator and denominator by $2\sqrt{5}+1$.

$$\frac{\left(3\sqrt{2}+1\right)\left(2\sqrt{5}+1\right)}{\left(2\sqrt{5}-1\right)\left(2\sqrt{5}+1\right)}$$

Consider $\left(2\sqrt{5}-1\right)\left(2\sqrt{5}+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(3\sqrt{2}+1\right)\left(2\sqrt{5}+1\right)}{\left(2\sqrt{5}\right)^{2}-1^{2}}$$

Expand $\left(2\sqrt{5}\right)^{2}$.

$$\frac{\left(3\sqrt{2}+1\right)\left(2\sqrt{5}+1\right)}{2^{2}\left(\sqrt{5}\right)^{2}-1^{2}}$$

Calculate $2$ to the power of $2$ and get $4$.

$$\frac{\left(3\sqrt{2}+1\right)\left(2\sqrt{5}+1\right)}{4\left(\sqrt{5}\right)^{2}-1^{2}}$$

The square of $\sqrt{5}$ is $5$.

$$\frac{\left(3\sqrt{2}+1\right)\left(2\sqrt{5}+1\right)}{4\times 5-1^{2}}$$

Multiply $4$ and $5$ to get $20$.

$$\frac{\left(3\sqrt{2}+1\right)\left(2\sqrt{5}+1\right)}{20-1^{2}}$$

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

$$\frac{\left(3\sqrt{2}+1\right)\left(2\sqrt{5}+1\right)}{20-1}$$

Subtract $1$ from $20$ to get $19$.

$$\frac{\left(3\sqrt{2}+1\right)\left(2\sqrt{5}+1\right)}{19}$$

Apply the distributive property by multiplying each term of $3\sqrt{2}+1$ by each term of $2\sqrt{5}+1$.

$$\frac{6\sqrt{2}\sqrt{5}+3\sqrt{2}+2\sqrt{5}+1}{19}$$

To multiply $\sqrt{2}$ and $\sqrt{5}$, multiply the numbers under the square root.

$$\frac{6\sqrt{10}+3\sqrt{2}+2\sqrt{5}+1}{19}$$

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