Question

If one zero of the polynomial (a^2+9)x^2+13x+6a is reciprocal of the other. Find a.

Answer 1

Sol:

Let the roots of (a² + 9) x² + 13x + 6a be p and 1/p.

Product of the roots = p x 1/p = (constant term) / coefficient of x²

⇒ (6a) / (a² + 9) = p x 1/p

⇒ (6a) / (a² + 9) = 1

⇒ a² - 6a + 9 = 0

⇒ (a - 3)² = 0

⇒ a - 3 = 0

⇒ a = 3

Therefore, the value of a is 3.