Question

The values of $p$ for which the equation $x^2-6|x|+5-|p|=0$ has exactly four real roots, is

• A. $(0,2)$
• B. $(-5,5)$
• C. $(2,10)$
• D. $(-5,10)$

Answer 1

The correct option is B $(-5,5)$

For four real roots, $t^2-6t+5-|p|=0(|x|=t)$ must have two distinct positive roots.
Let $f\left(t\right)=t^2-6t+5-|p|$
$\Rightarrow \Delta >0$ and $f\left(0\right)>0$
$\Rightarrow p\in R$ and $5-|p|>0$
$\Rightarrow p\in (-5,5)$