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Question

If one zero of polynomial (a​​​​​​2​​​​​+9)x​​​​​​2+13x+6a is reciprocal of other find value of a.

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Solution

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

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