Question

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

Answer 1

Find the value of $a$.

Given: One zero of $\left(a^2+9\right)x^2+13x+6a$ is reciprocal to the other.

let $\alpha$ be the one root of the quadratic equation. So, the other root will be $\frac{1}{\alpha }$.

We know that for the quadratic equation $a{x}^2+bx+c$ product of the roots is given by $\frac{c}{a}$.

Thus, product of the roots of the given quadratic equation $=\frac{6a}{a^2+9}$.

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

$$\begin{gathered} \Rightarrow \frac{6a}{a^2+9}=1 \\ \Rightarrow a^2-6a+9=0 \\ \Rightarrow {\left(a-3\right)}^2=0 \\ \Rightarrow a=3 \end{gathered}$$

Hence, the value of $a$ is $3$..