Range of k for roots of $|x^2-1|+x^2+kx=0$ lying between $(0,2)$

Question

If the equation $|x^2-1|+x^2+kx=0$ has $2$ distinct roots in $(0,2)$ then the range of $k$ is?

I tried this question as follows: If we open the mod with -ve sign then the quadratic is converted to linear which will not have $2$ solutions, so we open the mod in +ve sign and hence it becomes $2x^2+kx-1=0$. Now I used to nature of roots concept for roots between $(0,2)$ as it is upward parabola so $f(0)>0$, $f(2)>0$ these two conditions are necessary. But $f(0)=-1<0$ is a contradiction. Can you pls tell where i went wrong or is there any other way to solve?

Answer 1

Hint: Draw the graph of $ y = |x^2 - 1 | + x^2 $ on the domain $ (0, 2 )$.

From Wolfram,

Now, when would this graph intersect $ y = -kx $ at 2 distinct points?

Hence, conclude that $ k \in ( - \frac{7}{2} , -1 )$.

Answer 2

$|x^2 - 1| + x^2 + kx = kx + 1$ when $x\in [-1,1]$

If $k \ge -1$ we have one root in $(0,1]$

Now we need to find the range of $k$ such that.

$2x^2 + kx - 1$ has a root in $(1,2)$

$1 < \frac {-k + \sqrt {k^2 + 8}}{4}\\ 4+k < \sqrt {k^2 + 8}\\ 16 + 8k + k^2 < k^2 + 8\\ k<-1$

And

$2 > \frac {-k + \sqrt {k^2 + 8}}{4}\\ 8+k > \sqrt {k^2 + 8}\\ 64 + 16k + k^2 > k^2 + 8\\ 16k > -56\\ k > -\frac {7}{2}$

$k\in (-\frac {7}{2},-1)$