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

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Solution

We know that equation $a x^{2} + b x + c = 0$
Then sum of roots $= \frac{- b}{a}$ and product of roots $= \frac{c}{a}$
Let the other zero be $α$
Therefore, the other zero is $\frac{1}{α}$
Now, $α \times \frac{1}{α} = \frac{6 a}{a^{2} - 9}$
$=> 1 = \frac{6 a}{a^{2} - 9}$
$=> a^{2} + 9 - 6 a = 0$
$=> a^{2} - 6 a + 9 = 0$
$=> a^{2} - 3 a - 3 a + 9 = 0$
$=> a (a - 3) - 3 (a - 3) = 0$
$=> (a - 3) (a - 3) = 0$
$=> a = 3$ and $a = 3$

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