Question

If one zeros of the polynomial (x2+9)x2+3x+6a is reciprocal of the other,find the value of a

Answer 1

If one of the zero is reciprocal of the other, 

let one zero be α

other zero will be 1/α

Product of roots = c/a

⇒ α × 1/α = 6a/(a²+9)

⇒ 1 = 6a/(a²+9)

⇒ a² + 9 = 6a

⇒ a² - 6a + 9 = 0

⇒ a² - 3a - 3a + 9= 0

⇒ a(a-3) -3(a-3) = 0

⇒ (a-3)(a-3) = 0

⇒ (a-3)²= 0

⇒ a = 3

Value of a is 3.

Answer 2

let one zero of quadratic polynomial is b

Then other zero of quadratic polynomial is 1/b

product of zeros = b×1/b = 1

constant term / coefficient of x^2 = 1

or, 6a/a^2+9 = 1 ( note there should be a^2 instead of x^2 inside bracket )

or, 6a = a^2 +9

or, a^2 -6a+9 = 0

or, a^2-3a-3a +9 = 0

or, a(a-3)-3(a-3) = 0

or, (a-3)^2 = 0

or a = 3 .