Multiple choice

If equation $\displaystyle { px }^{ 2 }+qx+r=0$ has real roots, then $r=?$

$\displaystyle r\le \frac { 4p }{ { q }^{ 2 } } $
$\displaystyle r\ge \frac { { q }^{ 2 } }{ 4p } $
$\displaystyle r=\frac { p }{ { q }^{ 2 } } $
$\displaystyle r=\frac { { 4q }^{ 2 } }{ p } $

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