"Find the set of values of k for which |(x-4)(x+2)| = k has four solutions."
EDIT:
Ok so I thought I'd start with setting the modulus function equal to k and -k to get the two set of results.
Doing that I ended up with:
(x -4)(x +2) = k
and
(x – 4)(x +2) = -k .
After solving for x I got x = 4 or -2 OR (the other set of results) x = -4 and 2. -> not sure if my logic here is right, as I simply took the "negative" version of the function to get the second set of results…
I do not know how to progress with this question, the x-values don't really help me much here. Any help would be appreciated.
Posted this in the question so that it's easier to see 🙂
Best Answer
Solving the equation $(x-4)(x+2)=k$ gives you $x_{1,2}=1\pm\sqrt{9+k}$; solving the equation $(x-4)(x+2)=-k$ gives you $x_{3,4}=1\pm\sqrt{9-k}$.
So in order to get four solutions, the following conditions have to be met:
Therefore the solution is (given $k$ is an integer): $k\in\{-8,-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,7,8\}$