Question

If one zero of the polynomial (a² + 9)x² + 13x + 6a is the reciprocal of the other, find the value of a.

Answer 1

Given: Let the zeros be α and `1/α` for the polynomial (a² + 9)x² + 13x + 6a.

Step-wise calculation:

1. Product of zeros = `α xx (1/α) = 1`.

2. But product of zeros = `("Constant term")/("Coefficient of"  x^2)`

= `(6a)/(a^2 + 9)`

3. Therefore, `(6a)/(a^2 + 9) = 1`

⇒ 6a = a² + 9

4. Rearranging: a² – 6a + 9 = 0

⇒ (a – 3)² = 0

⇒ a = 3