Solve Find a and b if (3sqrt(3))/(sqrt(3) - sqrt(2)) = a + sqrt(b) | 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

$$\frac { 3 \sqrt { 3 } } { \sqrt { 3 } - \sqrt { 2 } } = a + \sqrt { b }$$

Solve for b

$b=\left(-a+3\sqrt{6}+9\right)^{2}$
$-a+3\sqrt{6}+9\geq 0$

Steps for Solving Radical Equations

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

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

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

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

Square $\sqrt{3}$. Square $\sqrt{2}$.

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

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

$$\frac{3\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{1}=a+\sqrt{b}$$

Anything divided by one gives itself.

$$3\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)=a+\sqrt{b}$$

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

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

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

$$3\times 3+3\sqrt{3}\sqrt{2}=a+\sqrt{b}$$

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

$$9+3\sqrt{3}\sqrt{2}=a+\sqrt{b}$$

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

$$9+3\sqrt{6}=a+\sqrt{b}$$

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

$$a+\sqrt{b}=9+3\sqrt{6}$$

Subtract $a$ from both sides.

$$\sqrt{b}=9+3\sqrt{6}-a$$

Square both sides of the equation.

$$b=\left(-a+3\sqrt{6}+9\right)^{2}$$

Solve for a (complex solution)

$a=-\sqrt{b}+3\sqrt{6}+9$

Solve for b (complex solution)

$b=\left(-a+3\sqrt{6}+9\right)^{2}$
$a=3\sqrt{6}+9\text{ or }arg(-a+3\sqrt{6}+9)<\pi $

Solve for a

$a=-\sqrt{b}+3\sqrt{6}+9$
$b\geq 0$