Suppose $a,b,c,p\in \mathbf{R}a\ne 0,c\ne 0$ and $b^2-4ac>0$

Statement-1: If the Roots of $f\left(x\right)=a{x}^2+bx+c=0$ are
symmetrically placed on the real line about the point p
then $p=-\frac{b}{2a}$ and $a\left(a{p}^2+bp+c\right)<0$
Statement-2: If the roots of $a{x}^2+bx+c=0$ are equal in magnitude but opposite in signs, then $b = 0$ and $ac < 0$.

If $\alpha ,-\alpha$ are roots of $a{x}^2+bx+c=0$

then $0=\alpha +(-\alpha )=-b/a\Rightarrow b=0$

Also $\alpha (-\alpha )=c/a$

As $c\ne 0,-\alpha \ne 0$ and therefore $c/a=-{\alpha }^2<0\Rightarrow ac<0$

Thus, Statement-2 is true.

If $p-\alpha$and $p+\alpha$ are roots of $f\left(x\right)=a{x}^2+bx+c=0$, then $\alpha ,-\alpha$ are roots of $f(x+p)=a(x+p{)}^2+b(x+p)+c=0$,

that is, $\alpha ,-\alpha$ are roots of $a{x}^2+(2ap+b)x+a{p}^2+bp+c=0$

By Statement-2:$2ap+b=0\Rightarrow p=-b/2a$ and $a\left(a{p}^2+bp+c\right)<0$.