If $\frac{1}{3-4i}$ is a root of $a{x}^2+bx+c=0,(a,b$

$c\in \mathbf{R},a\ne 0)$, then 

 TIP: As $a,b,c\in \mathbf{R}$ the other root must $\frac{1}{3+4i}$

Thus, $\frac{1}{3-4i}+\frac{1}{3+4i}=-\frac{b}{a},$

and $\frac{1}{3-4i}\cdot \frac{1}{3+4i}=\frac{c}{a}$

$\Rightarrow \frac{6}{25}=-\frac{b}{a}\text{ and }\frac{1}{25}=\frac{c}{a}$

Now, $6\left(\frac{c}{a}\right)=-\frac{b}{a}\Rightarrow b+6c=0$

Also, $a-25c=0$.