Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Form a quadratic polynomial whose zeroes are the reciprocals of the zeroes of the polynomial [tex]a x^2 + b x + c \ (a \neq 0)[/tex].

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]
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.