Question

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

Answer 1

Answer:

a = 3 .

Step-by-step explanation:

Answer 2

Answer:

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polynomial is (a2+9)x2 + 13x + 6a

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)

6a/(a2+9) = 1

⇒ 6a = a2+9

⇒ a2 -6a + 9 = 0

⇒ (a-3)2 = 0

⇒ a - 3 = 0

⇒ a = 3

polynomial will be (32+9)x2 + 13x + 6*3  

= 18x2 + 13x + 18

This polynomial will be have imaginary roots because b2-4ac<0