Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To form a quadratic polynomial whose zeroes are the reciprocals of the zeroes of the polynomial \( ax^2 + bx + c \), we need to follow a structured approach.
1. Given Polynomial:
The given quadratic polynomial is:
[tex]\[ f(x) = ax^2 + bx + c \][/tex]
2. Zeroes of the Given Polynomial:
Let's denote the zeroes of \( f(x) \) by \(\alpha\) and \(\beta\). Therefore, the polynomial can be written using its roots as:
[tex]\[ f(x) = a(x - \alpha)(x - \beta) \][/tex]
3. Reciprocal Zeroes:
We need to find a new polynomial whose roots are the reciprocals of \(\alpha\) and \(\beta\). Let's denote the reciprocal zeroes as \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).
4. Form of the New Polynomial:
The polynomial with roots \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) can be written as:
[tex]\[ g(x) = k(x - \frac{1}{\alpha})(x - \frac{1}{\beta}) \][/tex]
where \(k\) is a constant.
5. Expanding the New Polynomial:
Let's expand \(g(x)\):
[tex]\[ g(x) = k\left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right) \][/tex]
[tex]\[ g(x) = k \left[ x^2 - \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)x + \frac{1}{\alpha\beta} \right] \][/tex]
6. Relations Using Original Polynomial Coefficients:
From the properties of the roots of the polynomial \(ax^2 + bx + c\), we know:
- Sum of the roots \(\alpha + \beta = -\frac{b}{a}\)
- Product of the roots \(\alpha \beta = \frac{c}{a}\)
7. Substituting Relations:
Using these relations, we get:
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{c} \][/tex]
[tex]\[ \frac{1}{\alpha\beta} = \frac{1}{\frac{c}{a}} = \frac{a}{c} \][/tex]
8. Constructing the Polynomial:
Substitute these values into the polynomial:
[tex]\[ g(x) = k \left[ x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} \right] \][/tex]
[tex]\[ g(x) = k \left[ x^2 + \frac{b}{c}x + \frac{a}{c} \right] \][/tex]
9. Simplifying the Polynomial:
To match the standard form of a polynomial with integer coefficients, we choose \(k = c\). Therefore:
[tex]\[ g(x) = c \left[ x^2 + \frac{b}{c}x + \frac{a}{c} \right] \][/tex]
[tex]\[ g(x) = cx^2 + bx + a \][/tex]
Thus, the quadratic polynomial whose zeroes are the reciprocals of the zeroes of the polynomial \(ax^2 + bx + c \) is:
[tex]\[ g(x) = cx^2 + bx + a \][/tex]
1. Given Polynomial:
The given quadratic polynomial is:
[tex]\[ f(x) = ax^2 + bx + c \][/tex]
2. Zeroes of the Given Polynomial:
Let's denote the zeroes of \( f(x) \) by \(\alpha\) and \(\beta\). Therefore, the polynomial can be written using its roots as:
[tex]\[ f(x) = a(x - \alpha)(x - \beta) \][/tex]
3. Reciprocal Zeroes:
We need to find a new polynomial whose roots are the reciprocals of \(\alpha\) and \(\beta\). Let's denote the reciprocal zeroes as \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).
4. Form of the New Polynomial:
The polynomial with roots \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) can be written as:
[tex]\[ g(x) = k(x - \frac{1}{\alpha})(x - \frac{1}{\beta}) \][/tex]
where \(k\) is a constant.
5. Expanding the New Polynomial:
Let's expand \(g(x)\):
[tex]\[ g(x) = k\left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right) \][/tex]
[tex]\[ g(x) = k \left[ x^2 - \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)x + \frac{1}{\alpha\beta} \right] \][/tex]
6. Relations Using Original Polynomial Coefficients:
From the properties of the roots of the polynomial \(ax^2 + bx + c\), we know:
- Sum of the roots \(\alpha + \beta = -\frac{b}{a}\)
- Product of the roots \(\alpha \beta = \frac{c}{a}\)
7. Substituting Relations:
Using these relations, we get:
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{c} \][/tex]
[tex]\[ \frac{1}{\alpha\beta} = \frac{1}{\frac{c}{a}} = \frac{a}{c} \][/tex]
8. Constructing the Polynomial:
Substitute these values into the polynomial:
[tex]\[ g(x) = k \left[ x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} \right] \][/tex]
[tex]\[ g(x) = k \left[ x^2 + \frac{b}{c}x + \frac{a}{c} \right] \][/tex]
9. Simplifying the Polynomial:
To match the standard form of a polynomial with integer coefficients, we choose \(k = c\). Therefore:
[tex]\[ g(x) = c \left[ x^2 + \frac{b}{c}x + \frac{a}{c} \right] \][/tex]
[tex]\[ g(x) = cx^2 + bx + a \][/tex]
Thus, the quadratic polynomial whose zeroes are the reciprocals of the zeroes of the polynomial \(ax^2 + bx + c \) is:
[tex]\[ g(x) = cx^2 + bx + a \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.