Solve (2sqrt(3) - sqrt(5))/(2sqrt(2) + 3sqrt(3)) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Evaluate

$\frac{2\sqrt{10}+18-3\sqrt{15}-4\sqrt{6}}{19}\approx 0.153034016$

Solution Steps

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

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

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

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

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

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

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

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

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

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

Multiply $4$ and $2$ to get $8$.

$$\frac{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{2}-3\sqrt{3}\right)}{8-\left(3\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{2}-3\sqrt{3}\right)}{8-3^{2}\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{2}-3\sqrt{3}\right)}{8-9\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{2}-3\sqrt{3}\right)}{8-9\times 3}$$

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

$$\frac{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{2}-3\sqrt{3}\right)}{8-27}$$

Subtract $27$ from $8$ to get $-19$.

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

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

$$\frac{4\sqrt{3}\sqrt{2}-6\left(\sqrt{3}\right)^{2}-2\sqrt{5}\sqrt{2}+3\sqrt{3}\sqrt{5}}{-19}$$

To multiply $\sqrt{3}$ and $\sqrt{2}$, multiply the numbers under the square root.

$$\frac{4\sqrt{6}-6\left(\sqrt{3}\right)^{2}-2\sqrt{5}\sqrt{2}+3\sqrt{3}\sqrt{5}}{-19}$$

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

$$\frac{4\sqrt{6}-6\times 3-2\sqrt{5}\sqrt{2}+3\sqrt{3}\sqrt{5}}{-19}$$

Multiply $-6$ and $3$ to get $-18$.

$$\frac{4\sqrt{6}-18-2\sqrt{5}\sqrt{2}+3\sqrt{3}\sqrt{5}}{-19}$$

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

$$\frac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{3}\sqrt{5}}{-19}$$

To multiply $\sqrt{3}$ and $\sqrt{5}$, multiply the numbers under the square root.

$$\frac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{-19}$$

Multiply both numerator and denominator by -1.

$$\frac{-4\sqrt{6}+18+2\sqrt{10}-3\sqrt{15}}{19}$$