Question

If $b^2>4ac,$ then $a(x^2+4x+4{)}^2+b(x^2+4x+4)+c=0$ has distinct real roots if ?

• A. b < a < 0 < c
• B. a < b < 0 < c
• C. b < 0 < a < c
• D. c < 0 < b < a

Answer 1

The correct option is C b < 0 < a < c

Let $x^2+4x+4=y,$ then the equation becomes $a{y}^2+by+c=0$ ...(1)
$\because b^2-4ac>0\Rightarrow$ equation (1) has two distinct real roots, say $\alpha and\beta$
$\therefore y=\alpha or\beta \Rightarrow x^2+4x+4=\alpha or\beta$
$\Rightarrow (x+2{)}^2=\alpha or\beta$ ...(2)
Clearly equation (2) will give four distinct real values of x if $\alpha and\beta$ are positive. That is if equation (1) has positive roots. For this a and c should have the same sign and the sign of b should be negative. Only the option (c) satisfies this condition.