Is there any other ways to solve the domain of "a"?

Question

Assume $f(x)=|x^2 -5x| -a(x+4)$. If function $f(x)$ has just 4 roots , then what is the domain of real number $a$?

Is there any other ways to solve this? I mean the 2 gentlemen below use $\Delta$ to find the domain.

Answer 1

$\mid x^2-5x\mid=\begin {cases}5x-x^2,\,x\in[0,5]\\x^2-5x,\,x\notin[0,5]\end{cases}$.

You get two quadratic equations.

For $x\in[0,5]$, we get $f(x)=-x^2+(5-a)x-4a=0\implies f$ will have two roots if the discriminant is positive. $\Delta=(5-a)^2-4(4a)(1)=25-10a+a^2-16a=a^2-26a+25\gt0\implies a\notin [1,25]$.

I leave the other part for you.

Answer 2

1. Show that the equation $x^2-5x-a(x+4)=0$ has two solutions $ \iff a \in (- \infty,-25) \cup (-1, \infty).$

2. Show that the equation $-x^2+5x-a(x+4)=0$ has two solutions $ \iff a \in (- \infty,1) \cup (25, \infty).$

Can you take it from here ?