I've seen it before - 5

Since it has real roots, we have

$q^2-4pr\geq 0\rightarrow \text{Eq.1}$

since $p, q, r$ are in AP, we have

$\dfrac{p+r}{2}=q$

Substitute back this result in $\text{Eq.1}$ and solving further, we get

$p^2+r^2 \geq 14pr$

(Divide both sides by pr )

$\Rightarrow \dfrac{p}{r}+\dfrac{r}{p} \geq 14$

Let $\dfrac{p}{r}=a$

$a+\dfrac{1}{a}=14$

$\Rightarrow a^2-14a+1 \geq 0$

(Add 48 both sides)

$\Rightarrow a^2-14a+49 \geq 48$

$\Rightarrow (a-7)^2 \geq 48$

$\boxed{\Rightarrow \left|\dfrac{p}{r}-7\right| \geq 4\sqrt{3}}$

for real roots, discriminant of this equation must be $\ge$ 0 and as p,q,r are in arithmatic progression, so q= $\frac { p+r }{ 2 }$

now,

${ q }^{ 2 }-4pr\quad \ge 0\\ \Rightarrow { (\frac { p+r }{ 2 } ) }^{ 2 }\quad -4pr\quad \ge 0\\ \Rightarrow \quad { p }^{ 2 }-14pr+{ r }^{ 2 }\quad \ge 0\\ \Rightarrow \quad \frac { { p }^{ 2 } }{ { r }^{ 2 } } -14\frac { p }{ r } +\quad 1\quad \ge 0\\ \Rightarrow \quad \frac { { p }^{ 2 } }{ { r }^{ 2 } } -14\frac { p }{ r } \quad +\quad 49\quad \ge \quad 48\\ \Rightarrow \quad (\frac { p }{ r } -7)^{ 2 }\quad \ge \quad 48\\ \Rightarrow \quad \left| \frac { p }{ r } -7 \right| \quad \ge \quad 4\sqrt { 3 }$

(The problem was much easy to solve than latexing it :P )