For a $\ne$ 0 the equation ax² + b|x| + c = 0 has exactly ‘k’ real solutions and ‘p’ real roots
 

	COLUMN - I		COLUMN - II
a.	If k = 1, p = 2, then	p.	ab < 0
b.	If k = 2, p = 2, then	q.	ab=0
c.	If k = 3, then	r.	ac<0
d.	If k = 4, then	s.	ab>0
		t.	ac>0

${\mathrm{ax}}^2+\mathrm{b}\left|\mathrm{x}\right|+\mathrm{c}=0$ ; put $\left|\mathrm{x}\right|=\mathrm{t}\ge 0,\mathrm{\forall }\mathrm{x}\in \mathrm{R}$
$\therefore {\mathrm{at}}^2+\mathrm{bt}+\mathrm{c}=0$, t₁ and t₂ are roots of equation
a) for one solution : t₁ = 0 and t₂ = 0 $\therefore$ b = c = 0
$\Rightarrow$ ab = 0
b) for two solutions : t₁ < 0 and t₂ > 0
$\therefore$ product of roots $=\frac{\mathrm{c}}{\mathrm{a}}<0\Rightarrow \mathrm{ac}<0$
c) for three solution :t₁ = 0 and t₂ > 0
$\therefore$ sum of roots $=-\frac{\mathrm{b}}{\mathrm{a}}>0\Rightarrow \mathrm{ab}<0$
d) for four solutions : t₁ > 0 and t₂ > 0
$\therefore$ sum of roots $=-\frac{\mathrm{b}}{\mathrm{a}}>0$
and product of roots $=\frac{\mathrm{c}}{\mathrm{a}}>0$$\therefore$ ab < 0 and ac > 0