To solve for the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] given that [tex]\(\frac{a - \sqrt{48}}{\sqrt{3} + 1}\)[/tex] can be written in the form [tex]\(b \sqrt{3} - 9\)[/tex], follow these steps:
1. Rationalize the denominator of the given expression [tex]\(\frac{a - \sqrt{48}}{\sqrt{3} + 1}\)[/tex]:
Multiply the numerator and the denominator by the conjugate of the denominator [tex]\(\sqrt{3} - 1\)[/tex]:
[tex]\[
\frac{a - \sqrt{48}}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(a - \sqrt{48})(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}
\][/tex]
2. Simplify the denominator:
[tex]\[
(\sqrt{3} + 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2
\][/tex]
3. So, the expression becomes:
[tex]\[
\frac{(a - \sqrt{48})(\sqrt{3} - 1)}{2}
\][/tex]
4. Expand the numerator:
[tex]\[
(a - \sqrt{48})(\sqrt{3} - 1) = a\sqrt{3} - a - \sqrt{48}\sqrt{3} + \sqrt{48}
\][/tex]
Simplify [tex]\(\sqrt{48}\)[/tex]:
[tex]\[
\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}
\][/tex]
So the expression becomes:
[tex]\[
a\sqrt{3} - a - 4\sqrt{3}\sqrt{3} + 4\sqrt{3}
\][/tex]
Recall that [tex]\(\sqrt{3} \cdot \sqrt{3} = 3\)[/tex], thus:
[tex]\[
a\sqrt{3} - a - 4 \cdot 3 + 4\sqrt{3} = a\sqrt{3} - a - 12 + 4\sqrt{3}
\][/tex]
Combine like terms:
[tex]\[
(a + 4)\sqrt{3} - (a + 12)
\][/tex]
So now, the expression looks like:
[tex]\[
\frac{(a + 4)\sqrt{3} - (a + 12)}{2}
\][/tex]
5. Set the expression equal to the given form [tex]\(b \sqrt{3} - 9\)[/tex]:
[tex]\[
\frac{(a + 4)\sqrt{3} - (a + 12)}{2} = b \sqrt{3} - 9
\][/tex]
6. Separate and equate the coefficients of [tex]\(\sqrt{3}\)[/tex] and the constant terms:
[tex]\[
\frac{a + 4}{2} = b \quad \text{and} \quad \frac{-(a + 12)}{2} = -9
\][/tex]
Solve the second equation for [tex]\(a\)[/tex]:
[tex]\[
\frac{-(a + 12)}{2} = -9
\][/tex]
[tex]\[
-(a + 12) = -18
\][/tex]
[tex]\[
a + 12 = 18
\][/tex]
[tex]\[
a = 6
\][/tex]
7. Now find [tex]\(b\)[/tex] using the first equation:
[tex]\[
b = \frac{a + 4}{2} = \frac{6 + 4}{2} = \frac{10}{2} = 5
\][/tex]
Therefore, the values are [tex]\(a = 6\)[/tex] and [tex]\(b = 5\)[/tex].
The final answer is:
[tex]\[
\boxed{a = 6, \quad b = 5}
\][/tex]
