Solve (7+3sqrt(5))/(7-3sqrt(5))=a+sqrt(5b) | 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

$$\frac{ 7+3 \sqrt{ 5 } }{ 7-3 \sqrt{ 5 } } =a+ \sqrt{ 5b }$$

Solve for b

$b=\frac{\left(-\sqrt{5}\left(2a-47\right)+105\right)^{2}}{100}$
$-\sqrt{5}\left(\frac{a}{5}-\frac{47}{10}\right)+\frac{21}{2}\geq 0$

Steps for Solving Radical Equations

Rationalize the denominator of $\frac{7+3\sqrt{5}}{7-3\sqrt{5}}$ by multiplying numerator and denominator by $7+3\sqrt{5}$.

$$\frac{\left(7+3\sqrt{5}\right)\left(7+3\sqrt{5}\right)}{\left(7-3\sqrt{5}\right)\left(7+3\sqrt{5}\right)}=a+\sqrt{5b}$$

Consider $\left(7-3\sqrt{5}\right)\left(7+3\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(7+3\sqrt{5}\right)\left(7+3\sqrt{5}\right)}{7^{2}-\left(-3\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

Multiply $7+3\sqrt{5}$ and $7+3\sqrt{5}$ to get $\left(7+3\sqrt{5}\right)^{2}$.

$$\frac{\left(7+3\sqrt{5}\right)^{2}}{7^{2}-\left(-3\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(7+3\sqrt{5}\right)^{2}$.

$$\frac{49+42\sqrt{5}+9\left(\sqrt{5}\right)^{2}}{7^{2}-\left(-3\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

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

$$\frac{49+42\sqrt{5}+9\times 5}{7^{2}-\left(-3\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

Multiply $9$ and $5$ to get $45$.

$$\frac{49+42\sqrt{5}+45}{7^{2}-\left(-3\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

Add $49$ and $45$ to get $94$.

$$\frac{94+42\sqrt{5}}{7^{2}-\left(-3\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

Calculate $7$ to the power of $2$ and get $49$.

$$\frac{94+42\sqrt{5}}{49-\left(-3\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

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

$$\frac{94+42\sqrt{5}}{49-\left(-3\right)^{2}\left(\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

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

$$\frac{94+42\sqrt{5}}{49-9\left(\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

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

$$\frac{94+42\sqrt{5}}{49-9\times 5}=a+\sqrt{5b}$$

Multiply $9$ and $5$ to get $45$.

$$\frac{94+42\sqrt{5}}{49-45}=a+\sqrt{5b}$$

Subtract $45$ from $49$ to get $4$.

$$\frac{94+42\sqrt{5}}{4}=a+\sqrt{5b}$$

Divide each term of $94+42\sqrt{5}$ by $4$ to get $\frac{47}{2}+\frac{21}{2}\sqrt{5}$.

$$\frac{47}{2}+\frac{21}{2}\sqrt{5}=a+\sqrt{5b}$$

Swap sides so that all variable terms are on the left hand side.

$$a+\sqrt{5b}=\frac{47}{2}+\frac{21}{2}\sqrt{5}$$

Subtract $a$ from both sides.

$$\sqrt{5b}=\frac{47}{2}+\frac{21}{2}\sqrt{5}-a$$

Square both sides of the equation.

$$5b=\frac{\left(-2a+21\sqrt{5}+47\right)^{2}}{4}$$

Divide both sides by $5$.

$$\frac{5b}{5}=\frac{\left(-2a+21\sqrt{5}+47\right)^{2}}{4\times 5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$b=\frac{\left(-2a+21\sqrt{5}+47\right)^{2}}{4\times 5}$$

Divide $\frac{\left(21\sqrt{5}+47-2a\right)^{2}}{4}$ by $5$.

$$b=\frac{\left(-2a+21\sqrt{5}+47\right)^{2}}{20}$$

Solve for a

$a=-\sqrt{5b}+\frac{21\sqrt{5}}{2}+\frac{47}{2}$
$b\geq 0$