Answer :
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]
