If one root is 3 times of other of $a{x}^2+bx+c=0a\ne 0$ , then $16ac=$

Let 𝛼, 𝛽 are the roots of $a{x}^2+bx+c=0$; 𝛼 = 3𝛽  Sum of the roots

$\mathrm{Sum}=4\beta =\frac{-b}{a}$

$\Rightarrow {\beta }^2=\frac{b^2}{16a^2}\to \left(1\right)$

product $=3{\beta }^2=\frac{c}{a}$

$\Rightarrow {\beta }^2=\frac{c}{3a}\to \left(2\right)$

$\left(1\right) =\left(2\right)$

$\frac{b^2}{16a^2}=\frac{c}{3a}$

$16a^{2}c=3a{b}^2$

$\Rightarrow 16ac=3b^2$