Question

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

Answer 1

(a + 9)x² – 13x + 6a = 0 

Here, A = (a² + 9), B = 13 and C = 6a 

Let α and \(\frac{1}{\alpha}\) be the two zeroes. 

Then, product of the zeroes = \(\frac{C}A\)

⇒ α.\(\frac{1}{\alpha}\) = \(\frac{6a}{a^2+9}\)

⇒ 1 = \(\frac{6a}{a^2+9}\)

⇒ a² + 9 = 6a 

⇒ a² – 6a + 9 = 0 

⇒ a² – 2 × a × 3 + 3² = 0

⇒ (a – 3)² = 0 

⇒ a – 3 = 0 

⇒ a = 3