Polynomial Whose Zeroes Are Reciprocals of Given Zeroes

Video Explanation

Question

If \( \alpha \) and \( \beta \) are the zeroes of the polynomial

\[ f(x) = x^2 + px + q, \]

find the polynomial whose zeroes are

\[ \frac{1}{\alpha} \quad \text{and} \quad \frac{1}{\beta}. \]

Solution

Step 1: Write Relations Between Zeroes and Coefficients

For the polynomial \(x^2 + px + q\),

\[ \alpha + \beta = -p, \quad \alpha\beta = q \]

Step 2: Find Sum and Product of Reciprocal Zeroes

Sum of reciprocal zeroes:

\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-p}{q} \]

Product of reciprocal zeroes:

\[ \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{q} \]

Step 3: Form the Required Polynomial

The required quadratic polynomial is:

\[ x^2 – \left(\text{sum of zeroes}\right)x + \left(\text{product of zeroes}\right) \]

\[ = x^2 – \left(\frac{-p}{q}\right)x + \frac{1}{q} \]

\[ = x^2 + \frac{p}{q}x + \frac{1}{q} \]

Multiplying throughout by \(q\),

\[ qx^2 + px + 1 \]

Conclusion

The polynomial whose zeroes are \[ \frac{1}{\alpha} \text{ and } \frac{1}{\beta} \] is

\[ \boxed{qx^2 + px + 1} \]