Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
To solve the equation [tex]\(\frac{2 + 3 \sqrt{5}}{4 + 5 \sqrt{5}} = a + b \sqrt{5}\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are rational numbers, we need to rationalize the denominator.
Here's the step-by-step solution:
1. Given Equation:
[tex]\[ \frac{2+3\sqrt{5}}{4+5\sqrt{5}} \][/tex]
2. Rationalize the Denominator:
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(4 + 5\sqrt{5}\)[/tex] is [tex]\(4 - 5\sqrt{5}\)[/tex].
3. Multiply the Numerator and Denominator by the Conjugate:
[tex]\[ \frac{(2 + 3\sqrt{5})(4 - 5\sqrt{5})}{(4 + 5\sqrt{5})(4 - 5\sqrt{5})} \][/tex]
4. Simplify the Denominator:
Use the difference of squares formula: [tex]\((a + b)(a - b) = a^2 - b^2\)[/tex].
[tex]\[ (4 + 5\sqrt{5})(4 - 5\sqrt{5}) = 4^2 - (5\sqrt{5})^2 = 16 - 125 = -109 \][/tex]
5. Expand and Simplify the Numerator:
[tex]\[ (2 + 3\sqrt{5})(4 - 5\sqrt{5}) = 2 \cdot 4 + 2 \cdot (-5\sqrt{5}) + 3\sqrt{5} \cdot 4 + 3\sqrt{5} \cdot (-5\sqrt{5}) \][/tex]
Simplify each term:
[tex]\[ 2 \cdot 4 = 8, \quad 2 \cdot (-5\sqrt{5}) = -10\sqrt{5}, \quad 3\sqrt{5} \cdot 4 = 12\sqrt{5}, \quad 3\sqrt{5} \cdot (-5\sqrt{5}) = -75 \cdot 5 = -75 \][/tex]
Combine these terms:
[tex]\[ 8 - 75 - 10\sqrt{5} + 12\sqrt{5} \][/tex]
Simplify further:
[tex]\[ -67 + 2\sqrt{5} \][/tex]
6. Form the Fraction:
[tex]\[ \frac{-67 + 2\sqrt{5}}{-109} \][/tex]
7. Separate the Rational and Irrational Parts:
[tex]\[ \frac{-67}{-109} + \frac{2\sqrt{5}}{-109} \][/tex]
Simplify the fractions:
[tex]\[ \frac{-67}{-109} = \frac{67}{109} \quad \text{and} \quad \frac{2\sqrt{5}}{-109} = -\frac{2\sqrt{5}}{109} \][/tex]
Thus, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are:
[tex]\[ a = \frac{67}{109} \approx 0.6146788990825688, \quad b = -\frac{2}{109} \approx -0.01834862385321101 \][/tex]
Here's the step-by-step solution:
1. Given Equation:
[tex]\[ \frac{2+3\sqrt{5}}{4+5\sqrt{5}} \][/tex]
2. Rationalize the Denominator:
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\(4 + 5\sqrt{5}\)[/tex] is [tex]\(4 - 5\sqrt{5}\)[/tex].
3. Multiply the Numerator and Denominator by the Conjugate:
[tex]\[ \frac{(2 + 3\sqrt{5})(4 - 5\sqrt{5})}{(4 + 5\sqrt{5})(4 - 5\sqrt{5})} \][/tex]
4. Simplify the Denominator:
Use the difference of squares formula: [tex]\((a + b)(a - b) = a^2 - b^2\)[/tex].
[tex]\[ (4 + 5\sqrt{5})(4 - 5\sqrt{5}) = 4^2 - (5\sqrt{5})^2 = 16 - 125 = -109 \][/tex]
5. Expand and Simplify the Numerator:
[tex]\[ (2 + 3\sqrt{5})(4 - 5\sqrt{5}) = 2 \cdot 4 + 2 \cdot (-5\sqrt{5}) + 3\sqrt{5} \cdot 4 + 3\sqrt{5} \cdot (-5\sqrt{5}) \][/tex]
Simplify each term:
[tex]\[ 2 \cdot 4 = 8, \quad 2 \cdot (-5\sqrt{5}) = -10\sqrt{5}, \quad 3\sqrt{5} \cdot 4 = 12\sqrt{5}, \quad 3\sqrt{5} \cdot (-5\sqrt{5}) = -75 \cdot 5 = -75 \][/tex]
Combine these terms:
[tex]\[ 8 - 75 - 10\sqrt{5} + 12\sqrt{5} \][/tex]
Simplify further:
[tex]\[ -67 + 2\sqrt{5} \][/tex]
6. Form the Fraction:
[tex]\[ \frac{-67 + 2\sqrt{5}}{-109} \][/tex]
7. Separate the Rational and Irrational Parts:
[tex]\[ \frac{-67}{-109} + \frac{2\sqrt{5}}{-109} \][/tex]
Simplify the fractions:
[tex]\[ \frac{-67}{-109} = \frac{67}{109} \quad \text{and} \quad \frac{2\sqrt{5}}{-109} = -\frac{2\sqrt{5}}{109} \][/tex]
Thus, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are:
[tex]\[ a = \frac{67}{109} \approx 0.6146788990825688, \quad b = -\frac{2}{109} \approx -0.01834862385321101 \][/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.