Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
