Solve B = (1 - sqrt(18))(3 + sqrt(8)); C = (1 - sqrt(3))/(1 + sqrt(3)) + (1 + sqrt(3))/(1 - sqrt(3)) Exercice3 : | 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

$$\left. \begin{array} { l } { B = ( 1 - \sqrt { 18 } ) ( 3 + \sqrt { 8 } ) } \\ { C = \frac { 1 - \sqrt { 3 } } { 1 + \sqrt { 3 } } + \frac { 1 + \sqrt { 3 } } { 1 - \sqrt { 3 } } } \end{array} \right.$$

Solve for B, C

$B=-7\sqrt{2}-9\approx -18.899494937$
$C=-4$

Solution Steps

Consider the first equation. Factor $18=3^{2}\times 2$. Rewrite the square root of the product $\sqrt{3^{2}\times 2}$ as the product of square roots $\sqrt{3^{2}}\sqrt{2}$. Take the square root of $3^{2}$.

$$B=\left(1-3\sqrt{2}\right)\left(3+\sqrt{8}\right)$$

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 $2^{2}$.

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

Use the distributive property to multiply $1-3\sqrt{2}$ by $3+2\sqrt{2}$ and combine like terms.

$$B=3-7\sqrt{2}-6\left(\sqrt{2}\right)^{2}$$

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

$$B=3-7\sqrt{2}-6\times 2$$

Multiply $-6$ and $2$ to get $-12$.

$$B=3-7\sqrt{2}-12$$

Subtract $12$ from $3$ to get $-9$.

$$B=-9-7\sqrt{2}$$

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

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

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

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

Square $1$. Square $\sqrt{3}$.

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

Subtract $3$ from $1$ to get $-2$.

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

Multiply $1-\sqrt{3}$ and $1-\sqrt{3}$ to get $\left(1-\sqrt{3}\right)^{2}$.

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

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(1-\sqrt{3}\right)^{2}$.

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

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

$$C=\frac{1-2\sqrt{3}+3}{-2}+\frac{1+\sqrt{3}}{1-\sqrt{3}}$$

Add $1$ and $3$ to get $4$.

$$C=\frac{4-2\sqrt{3}}{-2}+\frac{1+\sqrt{3}}{1-\sqrt{3}}$$

Divide each term of $4-2\sqrt{3}$ by $-2$ to get $-2+\sqrt{3}$.

$$C=-2+\sqrt{3}+\frac{1+\sqrt{3}}{1-\sqrt{3}}$$

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

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

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

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

Square $1$. Square $\sqrt{3}$.

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

Subtract $3$ from $1$ to get $-2$.

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

Multiply $1+\sqrt{3}$ and $1+\sqrt{3}$ to get $\left(1+\sqrt{3}\right)^{2}$.

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

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(1+\sqrt{3}\right)^{2}$.

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

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

$$C=-2+\sqrt{3}+\frac{1+2\sqrt{3}+3}{-2}$$

Add $1$ and $3$ to get $4$.

$$C=-2+\sqrt{3}+\frac{4+2\sqrt{3}}{-2}$$

Divide each term of $4+2\sqrt{3}$ by $-2$ to get $-2-\sqrt{3}$.

$$C=-2+\sqrt{3}-2-\sqrt{3}$$

Subtract $2$ from $-2$ to get $-4$.

$$C=-4+\sqrt{3}-\sqrt{3}$$

Combine $\sqrt{3}$ and $-\sqrt{3}$ to get $0$.

$$C=-4$$

The system is now solved.

$$B=-9-7\sqrt{2}$$ $$C=-4$$