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

The product of zeros of a quadratic polynomial ax²+ bx + c is c/a.

Answer: If one zero of the polynomial (a² + 9)x² + 13x + 6a is the reciprocal of the other, then the value of a is 3.

Let us proceed step by step to find the value of a.

Explanation:

According to the question, one of the zeros is the reciprocal of the other, so, let us consider one zero to be x.

Therefore, the other zero will be 1 / x, and the product of zeros = x × 1/x = 1 ... (1)

For any polynomial of the form ax²+ bx + c = 0, the product of zeroes = c / a.

Using this, for the equation given in the question (a² + 9)x² + 13x + 6a, the product of zeros = 6a / (a²+ 9)

Then 6a / (a²+ 9) = 1 (From (1))

⇒ a²+ 9 = 6a

⇒ a²- 6a + 9 = 0 [rearranging terms]

Now, we will solve this quadratic equation by factoring.

⇒ a² – 3a – 3a + 9 = 0 [splitting middle term]

⇒ a (a - 3) - 3 (a - 3) = 0 [taking a as common in first two terms and - 3 as common in last two terms]

⇒ (a - 3) (a - 3) = 0

⇒ a = 3 , 3

Hence, if one zero of the polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other, then a = 3.