# If one zero of the polynomial (a^{2} + 9)x^{2} + 13x + 6a is reciprocal of the other, find the value of a.

A polynomial is defined as an algebraic expression whose maximum power of the variable must be a non-negative integer.

## Answer: If one zero of the polynomial (a^{2} + 9)x^{2} + 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 will be 1.

For any polynomial of the form ax^{2}+ bx + c = 0,

Sum of zeros = - b / a

Product of zeroes = c / a

Using these results for the equation given in the question (a^{2} + 9)x^{2} + 13x + 6a, we get

The product of zeros will be c / a = 6a / (a^{2}+ 9) = 1

⇒ a^{2}+ 9 = 6a

⇒ a^{2}- 6a + 9 = 0 [rearranging terms]

⇒ a^{2} – 3a – 3a + 9 = 0 [splitting middle term]

⇒ a (a - 3) -3 (a - 3) = 0 [taking a as common in 1st two terms and - 3 as common in last two terms]

⇒ (a - 3) (a - 3) = 0

⇒ a = 3 , 3

### Hence, the value of a is 3.

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