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,

$a{x}^2+(b-c)x+a-b-c=0$
Since, the roots are real and unequal.

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

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

$\Rightarrow c^2+(4a-2b)c-4a^2+4ab+b^2>0\text{for all}c\in R$

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

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

$\Rightarrow 4a^2-4ab+b^2+4a^2-4ab-b^2<0$

$\Rightarrow a(a-b)<0$

$\Rightarrow a<0$ and $a-b>0$ or $a>0$ and $a-b<0$

$\Rightarrow b<a<0$ or $b>a>0$