[tex]\displaystyle\bf\\9-\frac{1}{16} x^{2} =0\\\\\\3^{2} -\Big(\frac{1}{4} x\Big)^{2} =0\\\\\\\Big(3-\frac{1}{4} x\Big)\cdot\Big(3+\frac{1}{4} x\Big)=0\\\\\\\left[\begin{array}{ccc}3-\dfrac{1}{4} x=0\\3+\dfrac{1}{4} x=0\end{array}\right\\\\\\\left[\begin{array}{ccc}\dfrac{1}{4} x=3\\\dfrac{1}{4} x=-3\end{array}\right\\\\\\\left[\begin{array}{ccc} x_{1} =12\\ x_{2} =-12\end{array}\right\\\\\\x_{1} +x_{2} =12+(-12)=0[/tex]