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.
Best Answer
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.