Solve (3sqrt(3) - 2sqrt(7))/(2sqrt(3) + sqrt(7)) | 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

$$\frac{(3\sqrt{3}-2\sqrt{7})}{(2\sqrt{3}+\sqrt{7})}$$

Evaluate

$\frac{32-7\sqrt{21}}{5}\approx -0.015605973$

Solution Steps

Rationalize the denominator of $\frac{3\sqrt{3}-2\sqrt{7}}{2\sqrt{3}+\sqrt{7}}$ by multiplying numerator and denominator by $2\sqrt{3}-\sqrt{7}$.

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

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

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

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

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

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

$$\frac{\left(3\sqrt{3}-2\sqrt{7}\right)\left(2\sqrt{3}-\sqrt{7}\right)}{4\left(\sqrt{3}\right)^{2}-\left(\sqrt{7}\right)^{2}}$$

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

$$\frac{\left(3\sqrt{3}-2\sqrt{7}\right)\left(2\sqrt{3}-\sqrt{7}\right)}{4\times 3-\left(\sqrt{7}\right)^{2}}$$

Multiply $4$ and $3$ to get $12$.

$$\frac{\left(3\sqrt{3}-2\sqrt{7}\right)\left(2\sqrt{3}-\sqrt{7}\right)}{12-\left(\sqrt{7}\right)^{2}}$$

The square of $\sqrt{7}$ is $7$.

$$\frac{\left(3\sqrt{3}-2\sqrt{7}\right)\left(2\sqrt{3}-\sqrt{7}\right)}{12-7}$$

Subtract $7$ from $12$ to get $5$.

$$\frac{\left(3\sqrt{3}-2\sqrt{7}\right)\left(2\sqrt{3}-\sqrt{7}\right)}{5}$$

Apply the distributive property by multiplying each term of $3\sqrt{3}-2\sqrt{7}$ by each term of $2\sqrt{3}-\sqrt{7}$.

$$\frac{6\left(\sqrt{3}\right)^{2}-3\sqrt{3}\sqrt{7}-4\sqrt{3}\sqrt{7}+2\left(\sqrt{7}\right)^{2}}{5}$$

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

$$\frac{6\times 3-3\sqrt{3}\sqrt{7}-4\sqrt{3}\sqrt{7}+2\left(\sqrt{7}\right)^{2}}{5}$$

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

$$\frac{18-3\sqrt{3}\sqrt{7}-4\sqrt{3}\sqrt{7}+2\left(\sqrt{7}\right)^{2}}{5}$$

To multiply $\sqrt{3}$ and $\sqrt{7}$, multiply the numbers under the square root.

$$\frac{18-3\sqrt{21}-4\sqrt{3}\sqrt{7}+2\left(\sqrt{7}\right)^{2}}{5}$$

To multiply $\sqrt{3}$ and $\sqrt{7}$, multiply the numbers under the square root.

$$\frac{18-3\sqrt{21}-4\sqrt{21}+2\left(\sqrt{7}\right)^{2}}{5}$$

Combine $-3\sqrt{21}$ and $-4\sqrt{21}$ to get $-7\sqrt{21}$.

$$\frac{18-7\sqrt{21}+2\left(\sqrt{7}\right)^{2}}{5}$$

The square of $\sqrt{7}$ is $7$.

$$\frac{18-7\sqrt{21}+2\times 7}{5}$$

Multiply $2$ and $7$ to get $14$.

$$\frac{18-7\sqrt{21}+14}{5}$$

Add $18$ and $14$ to get $32$.

$$\frac{32-7\sqrt{21}}{5}$$