# If one zero of the polynomial (a^{2 }+ 2 ) x^{2} + 10 x +3a is the reciprocal of the other. Find a.

The zeroes of a polynomial is a value that makes the whole polynomial equal to zero. Sum of the zeroes is -b / a and product of zeroes is c / a.

## Answer: The value of a is 1, 2 for the polynomial (a^{2 }+ 2 ) x^{2} + 10 x +3a .

Let's find the value of a.

**Explanation:**

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

Let one of the zeros be α

The other zero will be 1/α (reciprocal of the first zero)

As we know that, for a given polynomial ax^{2} + bx + c = 0,

The product of zeroes = c / a, where c is the constant term a is the coefficient of x^{2} .

As per the question,

(a^{2 }+ 2 ) x^{2} + 10 x + 3a

Constant term = 3a, Coefficient of x^{2} = (a^{2 }+ 2 )

⇒ α × 1/ α = 3a / (a^{2 }+ 2 )

By solving the LHS and cross multiplying, we get

⇒ (a^{2 }+ 2 ) = 3a

⇒ a^{2 } - 3a + 2 = 0

⇒ a^{2 }- 2a - a + 2 = 0 (By splitting the middle term)

⇒ a (a - 2) -1 (a - 2) = 0

⇒ (a - 2) (a - 1) = 0

Thus, we have two values of a,

⇒ a = 2 and a = 1