If $$ \frac{6}{\sqrt{3}+3}=a+b \sqrt{3} $$, determine the values of $$ a $$ and $$ b $$.

Question

UpStudy AI · Accepted Answer

To express $$ \frac{6}{\sqrt{3} + 3} $$ in the form $$ a + b\sqrt{3} $$, follow these steps:

1. **Rationalize the denominator:**

Multiply the numerator and denominator by the conjugate of the denominator, which is $$ \sqrt{3} - 3 $$:

$$
   \frac{6}{\sqrt{3} + 3} \times \frac{\sqrt{3} - 3}{\sqrt{3} - 3} = \frac{6(\sqrt{3} - 3)}{(\sqrt{3})^2 - (3)^2}
   $$

2. **Simplify the denominator:**

$$
   (\sqrt{3})^2 - 3^2 = 3 - 9 = -6
   $$

So the expression becomes:

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

3. **Identify $$ a $$ and $$ b $$:**

Comparing with $$ a + b\sqrt{3} $$, we have:

$$
   a = 3 \quad \text{and} \quad b = -1
   $$

**Answer:** 
After simplifying, $$ a = 3 $$ and $$ b = -1 $$.
$$ a = 3 $$ and $$ b = -1 $$.

If \( \frac{6}{\sqrt{3}+3}=a+b \sqrt{3} \), determine the values of \( a \) and \( b \).