Question

If $a{x}^2+(b-c)x+a-b-c=0$ has unequal real roots for all $c\in R$, then

• A. $b$ $<$ $0$ $<$ $a$
• B. $a<$ $0$ $<$ $b$
• C. $b<$ $a$ $<0$
• D. $b>a>0$

Answer 1

Answer: (C), (D)

The correct options are
C $b<$ $a$ $<0$
D $b>a>0$

We have,

$b={(b-c)}^2-4a(a-b-c)>0$

or$b^2+c^2-2bc-4a^2+4ab+4ac>0$

or$c^2+(4a-2b)c-4a^2+4ab+b^2>0$

for all $c\in R$.

Discriminant of above expression in $c$ must be negative.

Hence,

${(4a-2b)}^2-4(-4a^2+4ab+b^2)<0$

or$4a^2-4ab+b^2+4a^2-4ab-b^2<0$

or$a(a-b)<0⟹a<0\&a-b>0$

or$a>0\&a-b<0⟹b<a<0$

or$b>a>0$