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If one zero of the polynomial $(a^{2} + 9) x^{2} + 13 x + 6 a$ is reciprocal of the other, find the value of a.

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3 and 3

-3 and 3

-3 and -3

3 and 4

answer is A.

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Detailed Solution

Given polynomial,

(a^{2} + 9) x^{2} + 13 x + 6 a

. The one zero of the given polynomial is reciprocal of other.
Let α be the one zero then

\frac{1}{α}

be the other.
We know that, in any equation

a x^{2} + bx + c = 0

Sum of zeroes = \frac{- b}{a}

and

Product of zeroes = \frac{c}{a}

Then,

α \times \frac{1}{α} = \frac{6 a}{a^{2} + 9}

\Rightarrow 1 = \frac{6 a}{a^{2} + 9}

\Rightarrow a^{2} + 9 = 6 a

\Rightarrow a^{2} - 6 a + 9 = 0

\Rightarrow a^{2} - 3 a - 3 a + 9 = 0

\Rightarrow a (a - 3) - 3 (a - 3) = 0

\Rightarrow (a - 3) (a - 3) = 0

\Rightarrow (a - 3) = 0 or (a - 3) = 0

\Rightarrow a = 3 or    a = 3

Hence, option (1) is correct.

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