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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]
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]
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