Question

If one zero of the polynomial $\left(a^2+2\right)x^2+10x+3a$ is the reciprocal of the other. find $a$.

Answer 1

Find the value of $a$.

Given: One zero of $\left(a^2+2\right)x^2+10x+3a$ 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, the product of the roots of the given quadratic equation$=\frac{3a}{a^2+2}$.

$\Rightarrow \frac{3a}{a^2+2}=\alpha \times \frac{1}{\alpha }$

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

$$\begin{gathered} \Rightarrow \left(a-2\right)=0 \\ \Rightarrow a=2 \end{gathered}$$

And

$$\begin{gathered} \Rightarrow \left(a-1\right)=0 \\ \Rightarrow a=1 \end{gathered}$$

Hence, the value of $a$ is $1$ and $2$.