Finding the set of values of k for which a modulus equation has exactly 4 roots In my assignment, I have the following question:

Find the set of values of k for which the equation $|x^2-1|+x=k$ has
exactly four roots.

What I've tried:
Removed the modulus and made two different equations $x^2+x-(k+1)=0$ and $x^2-x+(k-1)=0$. Calculated their discriminants such that $D_{1}$ and $D_{2}$ (the discriminants of the first and second equations, respectively) are both positive, I got $k∈(-5/4,5/4)$. After that, since in the first equation $x∈(-∞, -1]∪[1,∞)$, I used the quadratic formula and got $±√D_{1}∈(-∞,-1]∪[3,∞)$ and proceeding similarly for the second equation, I got $±√D_{2}∈(-1,3)$. I am not able to proceed any further, please help.
 A: Your two conditions should rather be
$$±√D_{1}∈(-∞,-1)∪(3,∞)$$
$$±√D_{2}∈(-1,3)$$
(to prevent $x^2-1$ from being equal to $0$).
The first one means
$$-√D_{1}∈(-∞,-1),\quad+√D_{1}∈(3,∞)$$
i.e. $$√D_{1}∈(1,∞)\cap(3,∞),$$i.e.
$D_1>9,$ i.e. $k>1.$ Similarly, the second one means $D_2<1,$
i.e. again $k>1.$
Conclusion: the set of values of $k$ for which $|x^2-1|+x=k$ has exactly four solutions is $(1,5/4).$
Another method was to plot $y=|x^2-1|+x$ and see which horizontal lines cut the curve four times.
A: 
Find the set of values of k for which the equation $|x^2-1|+x=k$ has  exactly four roots.


What I've tried:


Removed the modulus and made two different equations $x^2+x-(k+1)=0$ and $x^2-x+(k-1)=0$.

You have two cases

*

*$(x^2 - 1) \leq 0.$

*$(x^2 - 1) \geq 0.$
Note the overlap in the two cases, where $~x^2 - 1 = 0.~$  This overlap will be discussed following the discussion of the two individual cases.

$\underline{\text{Case 1} ~: x^2 - 1 \leq 0}$
Any roots of $x$ must be in the region $-1 \leq x \leq 1$ 
and the pertinent equation is 
$1 - x^2 + x = k \implies x^2 - x + (k-1) = 0 \implies $
$$x = \frac{1}{2} \left[1 \pm \sqrt{1 - 4(k-1)}~\right]
~=~ \frac{1}{2} \left[1 \pm \sqrt{5 - 4k}~\right]. \tag1 $$
To complete Case 1, you have to find all values of $k$ such that:

*

*$5 - 4k > 0$

*The corresponding roots represented in (1) above are both in the range $-1 \leq x \leq 1.$
For the moment, let $A$ denote $\sqrt{5 - 4k} \implies A > 0$.
Then, the two roots will be given by
$$\frac{1 + A}{2}, ~~\text{and}~~ \frac{1 - A}{2}. \tag2 $$
So, it is clear that both of the roots represented by (2) above will be in the range $~-1 \leq x \leq 1,~$ if and only if $A \leq 1.$
Therefore, you have that
$$0 < \sqrt{5 - 4k} \leq 1 \implies $$
$$0 < (5 - 4k) \leq 1 \implies 1 \leq k < \frac{5}{4}.$$

$\underline{\text{Case 2} ~: x^2 - 1 \geq 0}$
Any roots of $x$ must be in the region 
$x \leq -1~$ or $~x \geq 1$ 
and the pertinent equation is 
$x^2 - 1 + x = k \implies x^2 + x - (k+1) = 0 \implies $
$$x = \frac{1}{2} \left[-1 \pm \sqrt{1 + 4(k+1)}~\right]
~=~ \frac{1}{2} \left[-1 \pm \sqrt{5 + 4k}~\right]. \tag3 $$
To complete Case 2, you have to find all values of $k$ such that:

*

*$5 + 4k > 0$

*The corresponding roots represented in (3) above are both in the range $~x \leq -1~$ or $~x \geq 1.$
For the moment, let $A$ denote $\sqrt{5 + 4k} \implies A > 0$.
Then, the two roots will be given by
$$\frac{-1 + A}{2}, ~~\text{and}~~ \frac{-1 - A}{2}. \tag4 $$
So, it is clear that both of the roots represented by (4) above will be in the range  $~x \leq -1~$ or $~x \geq 1 ~$ if and only if $A \geq 3.$
Therefore, you have that
$$\sqrt{5 + 4k} \geq 3 \implies 5 + 4k \geq 9 \implies k \geq 1.$$

$\underline{\text{Final Analysis}}$
This portion of my response is intentionally going to be very slow and pedantic, because the target audience is Math students new to the topic.  For a more sophisticated audience, this response would have been significantly shorter.
The results so far are:

*

*The combined assumption that $1 \leq k <  \dfrac{5}{4}$ 
and $(x^2 - 1) \leq 0$, will yield two distinct roots that both satisfy $~(x^2 - 1) \leq 0.$


*The combined assumption that $1 \leq k $ 
and $(x^2 - 1) \geq 0$, will yield two distinct roots that both satisfy $~(x^2 - 1) \geq 0.$
So, superficially, you could wrongly infer that the correct answer is $1 \leq k < \frac{5}{4}.$  In order to get the correct answer, you have to consider the original question more carefully.
It is required to identify all values of $k$ that result in four distinct roots.  Case 1 identified a range for $k$ that would provide two distinct roots that satisfied $(x^2 - 1) \leq 0.$  Case 2 identified a range for $k$ that would provide two distinct roots that satisfied $(x^2 - 1) \geq 0.$
In order to complete this (very tricky) problem, you have to consider which values of $k$ will result in four distinct roots from Case 1 and Case 2 combined.
This will be done by:

*

*Changing the Case 1 specification to force $~(x^2 - 1) < 0,~$ and changing the Case 2 specification to force $~(x^2 - 1) > 0.$


*These changes will guarantee four distinct roots.


*Then, consider Case 1 and Case 2 separately.  In each case, manually identify what happens if a value of $k$ is permitted that results in at least one of the roots specifically satisfying $x^2 - 1 = 0.$
In Case 1, if the requirement had (instead) been that $(x^2 - 1) < 0$, then the resulting range would have been
$$1 < k < \frac{5}{4}.$$
In Case 2, if the requirement had (instead) been that $(x^2 - 1) > 0$, then the resulting range would have been
$$1 < k.$$
Therefore, before deciding on a final answer, it remains to consider exactly what happens when $k = 1.$
Consider the equation
$$|x^2 - 1| + x = 1.$$
If the Case 1 assumption is made that $x^2 - 1 \leq 0$, then the two roots are $0$ and $1$.
If the Case 2 assumption is made that $x^2 - 1 \geq 0$, then the two roots are $-2$ and $1$.
So, the specific value of $k = 1$ generates only three distinct roots, rather than four.
Therefore, the final answer is
$$1 < k < \frac{5}{4}.$$
