Question

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

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

Answer 1

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

Given: $x^2-6|x|+5-|p|=0...\left(1\right)$

Let $|x|=t,$ then equation becomes:
$t^2-6t+5-|p|=0...\left(2\right)$
For the eq.$\left(1\right)$, to have four distinct real roots, eq.$\left(2\right)$ must have two distinct positive roots.

So, $t^2-6t+5-|p|=0$ should have both roots greater than zero.
Let $f\left(t\right)=t^2-6t+5-|p|$

Using the concept of location of roots, we have
$\left(i\right)D>0\left(ii\right)-\frac{b}{2a}>0\left(iii\right)f\left(0\right)>0$


$\left(i\right)D>0\Rightarrow (-6{)}^2+4|p|>0\Rightarrow 36+4|p|>0\Rightarrow p\in R$

$\left(ii\right)-\frac{b}{2a}>0\Rightarrow -\frac{(-6)}{2}>0\Rightarrow p\in R$

$\left(iii\right)f\left(0\right)>0\Rightarrow 5-|p|>0\Rightarrow 5>|p|\Rightarrow |p|<5\Rightarrow p\in (-5,5)$

From cases $\left(i\right),\left(ii\right),\left(iii\right)$ $\Rightarrow p\in (-5,5)$