Question

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

Answer 1

Let one zero of the given polynomial is α.

Given that one zero of given polynomial is reciprocal of the other.

∴ The other zero of given polynomial is \(\frac{1}{a}.\)

Hence, the zeros of the given polynomial (α² + 9)x² + 13x + 6 are and \(\frac{1}{a}.\)

Now, the product of the zeros is α. \(\frac{1}{a}\) = \(\frac{6a}{a^2+9}\)

⇒ \(\frac{6a}{a^2+9}=1\)

⇒ α² + 9 = 6a

⇒ α² − 6 + 9 = 0

⇒ (α − 3)² = 0 (∵ (α − b)² = α² − 2ab + b²)

⇒ α = 3.

Hence, the value of is 3.