Question:
If one zero of the polynomial $\left(a^{2}+9\right) x^{2}+13 x+6 a$ is the reciprocal of the other, find the value of $a$.
Solution:
$(a+9) x^{2}-13 x+6 a=0$
Here, $A=\left(a^{2}+9\right), B=13$ and $C=6 a$
Let $\alpha$ and $\frac{1}{\alpha}$ be the two zeroes.
Then, product of the zeroes $=\frac{C}{A}$
$=>\alpha \cdot \frac{1}{\alpha}=\frac{6 a}{a^{2}+9}$
$=>1=\frac{6 a}{a^{2}+9}$
$=>a^{2}+9=6 a$
$=>a^{2}-6 a+9=0$
$=>a^{2}-2 \times a \times 3+3^{2}=0$
$=>(a-3)^{2}=0$
$=>a-3=0$
$=>a=3$
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