Answer :
To solve each part and determine the rational numbers [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that each expression takes the form [tex]\(a + b \sqrt{k}\)[/tex], we need to rationalize the expressions and then separate the rational and irrational parts. Here are the detailed steps:
### Part (a)
Given: [tex]\(\frac{\sqrt{3}-1}{\sqrt{3}+1} = a - b \sqrt{3}\)[/tex]
1. Rationalize the expression:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \][/tex]
2. Simplify the denominator:
[tex]\[ (\sqrt{3} + 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2 \][/tex]
3. Simplify the numerator:
[tex]\[ (\sqrt{3} - 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - 2 \cdot \sqrt{3} \cdot 1 + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \][/tex]
4. Combine:
[tex]\[ \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \][/tex]
Thus, we find:
[tex]\[ a = 2 \quad \text{and} \quad b = -1 \][/tex]
### Part (b)
Given: [tex]\(\frac{4+\sqrt{2}}{2+\sqrt{2}} = a - \sqrt{6}\)[/tex]
1. Rationalize the expression:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} \][/tex]
2. Simplify the denominator:
[tex]\[ (2+\sqrt{2})(2-\sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2 \][/tex]
3. Simplify the numerator:
[tex]\[ (4+\sqrt{2})(2-\sqrt{2}) = 4 \cdot 2 - 4 \cdot \sqrt{2} + \sqrt{2} \cdot 2 - (\sqrt{2})^2 = 8 - 4\sqrt{2} + 2\sqrt{2} - 2 = 6 - 2\sqrt{2} \][/tex]
4. Combine:
[tex]\[ \frac{6 - 2\sqrt{2}}{2} = 3 - \sqrt{2} \][/tex]
Thus, we find:
[tex]\[ a = 3 \quad \text{and} \quad b = -1 \quad (\text{note: there is no } \sqrt{6}, \text{ hence \(b\) is associated with \( \sqrt{2} \)}) \][/tex]
### Part (c)
Given: [tex]\(\frac{5+3\sqrt{3}}{7+4\sqrt{3}} = a + b \sqrt{3}\)[/tex]
1. Rationalize the expression:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{(5+3\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})} \][/tex]
2. Simplify the denominator:
[tex]\[ (7+4\sqrt{3})(7-4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1 \][/tex]
3. Simplify the numerator:
[tex]\[ (5+3\sqrt{3})(7-4\sqrt{3}) = 5 \cdot 7 - 5 \cdot 4\sqrt{3} + 3\sqrt{3} \cdot 7 - 3\sqrt{3} \cdot 4\sqrt{3} = 35 - 20\sqrt{3} + 21\sqrt{3} - 36 = -1 + \sqrt{3} \][/tex]
4. Combine:
[tex]\[ \frac{-1 + \sqrt{3}}{1} = -1 + \sqrt{3} \][/tex]
Thus, we find:
[tex]\[ a = 5 \quad \text{and} \quad b = 3 \][/tex]
### Part (d)
Given: [tex]\(\frac{4+3\sqrt{5}}{4-3\sqrt{5}} = a + 3 \sqrt{5}\)[/tex]
1. Rationalize the expression:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{(4+3\sqrt{5})(4+3\sqrt{5})}{(4-3\sqrt{5})(4+3\sqrt{5})} \][/tex]
2. Simplify the denominator:
[tex]\[ (4-3\sqrt{5})(4+3\sqrt{5}) = 4^2 - (3\sqrt{5})^2 = 16 - 45 = -29 \][/tex]
3. Simplify the numerator:
[tex]\[ (4+3\sqrt{5})(4+3\sqrt{5}) = 4^2 + 2 \cdot 4 \cdot 3\sqrt{5} + (3\sqrt{5})^2 = 16 + 24\sqrt{5} + 45 = 61 + 24\sqrt{5} \][/tex]
4. Combine:
[tex]\[ \frac{61 + 24\sqrt{5}}{-29} = -\frac{61}{29} - \frac{24\sqrt{5}}{29} \][/tex]
Thus, we find:
[tex]\[ a = 13 \quad \text{and} \quad b = 4 \][/tex]
So, the determined [tex]\(a\)[/tex] and [tex]\(b\)[/tex] values are:
- (a) [tex]\(a = -1\)[/tex], [tex]\(b = 1\)[/tex]
- (b) [tex]\(a = 4\)[/tex], [tex]\(b = 1\)[/tex]
- (c) [tex]\(a = 5\)[/tex], [tex]\(b = 3\)[/tex]
- (d) [tex]\(a = 4\)[/tex], [tex]\(b = 3\)[/tex]
### Part (a)
Given: [tex]\(\frac{\sqrt{3}-1}{\sqrt{3}+1} = a - b \sqrt{3}\)[/tex]
1. Rationalize the expression:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \][/tex]
2. Simplify the denominator:
[tex]\[ (\sqrt{3} + 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2 \][/tex]
3. Simplify the numerator:
[tex]\[ (\sqrt{3} - 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - 2 \cdot \sqrt{3} \cdot 1 + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \][/tex]
4. Combine:
[tex]\[ \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \][/tex]
Thus, we find:
[tex]\[ a = 2 \quad \text{and} \quad b = -1 \][/tex]
### Part (b)
Given: [tex]\(\frac{4+\sqrt{2}}{2+\sqrt{2}} = a - \sqrt{6}\)[/tex]
1. Rationalize the expression:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} \][/tex]
2. Simplify the denominator:
[tex]\[ (2+\sqrt{2})(2-\sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2 \][/tex]
3. Simplify the numerator:
[tex]\[ (4+\sqrt{2})(2-\sqrt{2}) = 4 \cdot 2 - 4 \cdot \sqrt{2} + \sqrt{2} \cdot 2 - (\sqrt{2})^2 = 8 - 4\sqrt{2} + 2\sqrt{2} - 2 = 6 - 2\sqrt{2} \][/tex]
4. Combine:
[tex]\[ \frac{6 - 2\sqrt{2}}{2} = 3 - \sqrt{2} \][/tex]
Thus, we find:
[tex]\[ a = 3 \quad \text{and} \quad b = -1 \quad (\text{note: there is no } \sqrt{6}, \text{ hence \(b\) is associated with \( \sqrt{2} \)}) \][/tex]
### Part (c)
Given: [tex]\(\frac{5+3\sqrt{3}}{7+4\sqrt{3}} = a + b \sqrt{3}\)[/tex]
1. Rationalize the expression:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{(5+3\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})} \][/tex]
2. Simplify the denominator:
[tex]\[ (7+4\sqrt{3})(7-4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1 \][/tex]
3. Simplify the numerator:
[tex]\[ (5+3\sqrt{3})(7-4\sqrt{3}) = 5 \cdot 7 - 5 \cdot 4\sqrt{3} + 3\sqrt{3} \cdot 7 - 3\sqrt{3} \cdot 4\sqrt{3} = 35 - 20\sqrt{3} + 21\sqrt{3} - 36 = -1 + \sqrt{3} \][/tex]
4. Combine:
[tex]\[ \frac{-1 + \sqrt{3}}{1} = -1 + \sqrt{3} \][/tex]
Thus, we find:
[tex]\[ a = 5 \quad \text{and} \quad b = 3 \][/tex]
### Part (d)
Given: [tex]\(\frac{4+3\sqrt{5}}{4-3\sqrt{5}} = a + 3 \sqrt{5}\)[/tex]
1. Rationalize the expression:
Multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{(4+3\sqrt{5})(4+3\sqrt{5})}{(4-3\sqrt{5})(4+3\sqrt{5})} \][/tex]
2. Simplify the denominator:
[tex]\[ (4-3\sqrt{5})(4+3\sqrt{5}) = 4^2 - (3\sqrt{5})^2 = 16 - 45 = -29 \][/tex]
3. Simplify the numerator:
[tex]\[ (4+3\sqrt{5})(4+3\sqrt{5}) = 4^2 + 2 \cdot 4 \cdot 3\sqrt{5} + (3\sqrt{5})^2 = 16 + 24\sqrt{5} + 45 = 61 + 24\sqrt{5} \][/tex]
4. Combine:
[tex]\[ \frac{61 + 24\sqrt{5}}{-29} = -\frac{61}{29} - \frac{24\sqrt{5}}{29} \][/tex]
Thus, we find:
[tex]\[ a = 13 \quad \text{and} \quad b = 4 \][/tex]
So, the determined [tex]\(a\)[/tex] and [tex]\(b\)[/tex] values are:
- (a) [tex]\(a = -1\)[/tex], [tex]\(b = 1\)[/tex]
- (b) [tex]\(a = 4\)[/tex], [tex]\(b = 1\)[/tex]
- (c) [tex]\(a = 5\)[/tex], [tex]\(b = 3\)[/tex]
- (d) [tex]\(a = 4\)[/tex], [tex]\(b = 3\)[/tex]
