Question

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

Answer 1

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

Or a=3

Answer 2

Answer:

Let one zero be b then other zero will be reciprocal of it i.e.1/b.

Let one zero be b then other zero will be reciprocal of it i.e.1/b.∴ product of the zeroes = constant term/cofficient of x²= 1 (as b*1/b = 1)

of x²= 1 (as b*1/b = 1)6a/(a²+9) = 1

of x²= 1 (as b*1/b = 1)6a/(a²+9) = 1⇒ 6a = a²+9

of x²= 1 (as b*1/b = 1)6a/(a²+9) = 1⇒ 6a = a²+9⇒ a²-6a + 9 = 0

of x²= 1 (as b*1/b = 1)6a/(a²+9) = 1⇒ 6a = a²+9⇒ a²-6a + 9 = 0⇒ (a-3)² = 0

⇒ a - 3 = 0

a - 3 = 0⇒ a = 3

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