Solve ((6 + sqrt(3))/(sqrt(8) - sqrt(3)))((6 - sqrt(3))/(sqrt(8) + sqrt(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

$$(\frac{6+\sqrt{3}}{\sqrt{8}-\sqrt{3}})(\frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}})$$

Evaluate

$\frac{33}{5}=6.6$

Solution Steps

Factor $8=2^{2}\times 2$. Rewrite the square root of the product $\sqrt{2^{2}\times 2}$ as the product of square roots $\sqrt{2^{2}}\sqrt{2}$. Take the square root of $2^{2}$.

$$\frac{6+\sqrt{3}}{2\sqrt{2}-\sqrt{3}}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

Rationalize the denominator of $\frac{6+\sqrt{3}}{2\sqrt{2}-\sqrt{3}}$ by multiplying numerator and denominator by $2\sqrt{2}+\sqrt{3}$.

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

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

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

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

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{2^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

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

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{4\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

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

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{4\times 2-\left(\sqrt{3}\right)^{2}}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

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

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{8-\left(\sqrt{3}\right)^{2}}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

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

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{8-3}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

Subtract $3$ from $8$ to get $5$.

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{5}\times \frac{6-\sqrt{3}}{\sqrt{8}+\sqrt{3}}$$

Factor $8=2^{2}\times 2$. Rewrite the square root of the product $\sqrt{2^{2}\times 2}$ as the product of square roots $\sqrt{2^{2}}\sqrt{2}$. Take the square root of $2^{2}$.

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

Rationalize the denominator of $\frac{6-\sqrt{3}}{2\sqrt{2}+\sqrt{3}}$ by multiplying numerator and denominator by $2\sqrt{2}-\sqrt{3}$.

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

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

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

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

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

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

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

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

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

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

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{5}\times \frac{\left(6-\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{8-\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{5}\times \frac{\left(6-\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{8-3}$$

Subtract $3$ from $8$ to get $5$.

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

Multiply $\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{5}$ times $\frac{\left(6-\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{5}$ by multiplying numerator times numerator and denominator times denominator.

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

Multiply $5$ and $5$ to get $25$.

$$\frac{\left(6+\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)\left(6-\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Apply the distributive property by multiplying each term of $6+\sqrt{3}$ by each term of $2\sqrt{2}+\sqrt{3}$.

$$\frac{\left(12\sqrt{2}+6\sqrt{3}+2\sqrt{3}\sqrt{2}+\left(\sqrt{3}\right)^{2}\right)\left(6-\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

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

$$\frac{\left(12\sqrt{2}+6\sqrt{3}+2\sqrt{6}+\left(\sqrt{3}\right)^{2}\right)\left(6-\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

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

$$\frac{\left(12\sqrt{2}+6\sqrt{3}+2\sqrt{6}+3\right)\left(6-\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Apply the distributive property by multiplying each term of $12\sqrt{2}+6\sqrt{3}+2\sqrt{6}+3$ by each term of $6-\sqrt{3}$.

$$\frac{\left(72\sqrt{2}-12\sqrt{3}\sqrt{2}+36\sqrt{3}-6\left(\sqrt{3}\right)^{2}+12\sqrt{6}-2\sqrt{3}\sqrt{6}+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

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

$$\frac{\left(72\sqrt{2}-12\sqrt{6}+36\sqrt{3}-6\left(\sqrt{3}\right)^{2}+12\sqrt{6}-2\sqrt{3}\sqrt{6}+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

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

$$\frac{\left(72\sqrt{2}-12\sqrt{6}+36\sqrt{3}-6\times 3+12\sqrt{6}-2\sqrt{3}\sqrt{6}+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

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

$$\frac{\left(72\sqrt{2}-12\sqrt{6}+36\sqrt{3}-18+12\sqrt{6}-2\sqrt{3}\sqrt{6}+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Combine $-12\sqrt{6}$ and $12\sqrt{6}$ to get $0$.

$$\frac{\left(72\sqrt{2}+36\sqrt{3}-18-2\sqrt{3}\sqrt{6}+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Factor $6=3\times 2$. Rewrite the square root of the product $\sqrt{3\times 2}$ as the product of square roots $\sqrt{3}\sqrt{2}$.

$$\frac{\left(72\sqrt{2}+36\sqrt{3}-18-2\sqrt{3}\sqrt{3}\sqrt{2}+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Multiply $\sqrt{3}$ and $\sqrt{3}$ to get $3$.

$$\frac{\left(72\sqrt{2}+36\sqrt{3}-18-2\times 3\sqrt{2}+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Multiply $-2$ and $3$ to get $-6$.

$$\frac{\left(72\sqrt{2}+36\sqrt{3}-18-6\sqrt{2}+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Combine $72\sqrt{2}$ and $-6\sqrt{2}$ to get $66\sqrt{2}$.

$$\frac{\left(66\sqrt{2}+36\sqrt{3}-18+18-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Add $-18$ and $18$ to get $0$.

$$\frac{\left(66\sqrt{2}+36\sqrt{3}-3\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Combine $36\sqrt{3}$ and $-3\sqrt{3}$ to get $33\sqrt{3}$.

$$\frac{\left(66\sqrt{2}+33\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}{25}$$

Apply the distributive property by multiplying each term of $66\sqrt{2}+33\sqrt{3}$ by each term of $2\sqrt{2}-\sqrt{3}$.

$$\frac{132\left(\sqrt{2}\right)^{2}-66\sqrt{3}\sqrt{2}+66\sqrt{3}\sqrt{2}-33\left(\sqrt{3}\right)^{2}}{25}$$

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

$$\frac{132\times 2-66\sqrt{3}\sqrt{2}+66\sqrt{3}\sqrt{2}-33\left(\sqrt{3}\right)^{2}}{25}$$

Multiply $132$ and $2$ to get $264$.

$$\frac{264-66\sqrt{3}\sqrt{2}+66\sqrt{3}\sqrt{2}-33\left(\sqrt{3}\right)^{2}}{25}$$

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

$$\frac{264-66\sqrt{6}+66\sqrt{3}\sqrt{2}-33\left(\sqrt{3}\right)^{2}}{25}$$

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

$$\frac{264-66\sqrt{6}+66\sqrt{6}-33\left(\sqrt{3}\right)^{2}}{25}$$

Combine $-66\sqrt{6}$ and $66\sqrt{6}$ to get $0$.

$$\frac{264-33\left(\sqrt{3}\right)^{2}}{25}$$

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

$$\frac{264-33\times 3}{25}$$

Multiply $-33$ and $3$ to get $-99$.

$$\frac{264-99}{25}$$

Subtract $99$ from $264$ to get $165$.

$$\frac{165}{25}$$

Reduce the fraction $\frac{165}{25}$ to lowest terms by extracting and canceling out $5$.

$$\frac{33}{5}$$

Factor

$\frac{3 \cdot 11}{5} = 6\frac{3}{5} = 6.6$