$p+((2p-3)x+3p-1)$$|x|$$=2$
for which P does this equation have 1 real solution.
so, at first, I opened brackets and I get $(2p-3)|x|*x+(3p-1)|x|+p-2=0$ next I try to analyze $x$ $\geq$ $0$ and $x<0$ situations for these equations, but then I have to analyze $x \geq 0$ has 1 real solution and $x<0$ has no real solution, then there can be $x\geq0$ and $D=0$ and $x<0$ has no solution, then $X<0$ $D=0$ and $x\geq0$ has no solution, I am asking if there are any generalized inequalities than can help me to solve this equation, without analyzing separate cases too much.
This can look like a very tangled set of situations to consider; studying the behavior of the curve for the function can be helpful in understanding how to analyze it.
A way to think about the equation $ \ (2p-3)x|x| \ + \ (3p-1)|x| \ + \ (p - 2) \ = \ 0 \ $ is to consider the left side as the description of pieces of two parabolas "glued together" at the $ \ y-$axis,
• for $ \ x \ \ge \ 0 \ \ , \ \ y \ = \ (2p - 3)·x^2 \ + \ (3p - 1)·x \ + \ (p - 2) \ \ ; $
• for $ \ x \ \le \ 0 \ \ , \ \ y \ = \ (3 - 2p)·x^2 \ + \ (1 - 3p)·x \ + \ (p - 2) \ \ . $
We need to consider the zeroes of each parabola, which correspond to intersections with the line $ \ y \ = \ 2 \ \ . $
The "right-side $ \ ( \ x \ge 0 \ ) \ $ parabola" has its $ y-$intercept at $ \ (0 \ , \ p - 2) \ $ , its vertex at $ \ x_{v} \ = \ \frac{1 \ - \ 3p}{2·(2p \ - \ 3)} \ , $ and the discriminant of the polynomial is $ \ \Delta \ = \ p^2 + 22p - 23 \ = \ (p + 23)·(p - 1) \ $ ; it "opens downward" for $ \ p \ < \ \frac32 \ $ and "upward" for $ \ p \ > \ \frac32 \ \ . $ The vertex is only on or "to the right" of the $ \ y-$axis for $ \ \frac13 \ \le \ p \ < \ \frac32 \ \ . $ Putting this information together tells us that the parabola only intersects the $ \ x-$axis for $ \ x \ > \ 0 \ $ with $ \ p \ \ge \ 1 \ \ , $ becomes the straight line $ \ \frac72·x - \frac12 \ $ for $ \ p \ = \ \frac32 \ \ $ (with $ \ x-$intercept $ \ x \ = \ \frac17 \ ) \ , $ and "opens upward" with $ \ y-$intercept $ \ y \ < \ 0 \ \ $ up to $ \ p \ = \ 2 \ \ ; $ there are no $ x-$intercepts with $ \ x \ \ge \ 0 \ $ for $ \ p \ > \ 2 \ \ . $
To summarize for $ \ x \ \ge \ 0 \ $ then, there are no intersections with $ \ y \ = \ 2 \ $ for $ \ p \ < \ 1 \ \ , $ one intersection for $ \ p \ = \ 1 \ \ , $ two intersections for $ \ 1 \ < \ p \ < \ \frac32 \ \ , $ one for $ \ \frac32 \ \le \ p \ \le \ 2 \ \ . $ and none for $ \ p \ > \ 2 \ \ . $
We continue in the same vein for the "left-side $ \ ( \ x \le 0 \ ) \ $ parabola". Its $ y-$intercept is at $ \ (0 \ , \ p - 2) \ $ and its vertex at $ \ x_{v} \ = \ \frac{1 \ - \ 3p}{2·(2p \ - \ 3)} \ $ again, with the discriminant now $ \ \Delta \ = \ 17p^2 - 34p + 25 \ = \ 17·(p - 1)^2 + 8 \ \ , $ which is always positive. This parabola "opens upward" for $ \ p \ < \ \frac32 \ $ and "downward" for $ \ p \ > \ \frac32 \ \ . $ The vertex is "to the left" of the $ \ y-$axis for $ \ p \ < \ \frac13 \ $ or $ \ p \ > \ \frac32 \ \ . $ What happens for this parabola is that because the $ \ y-$intercept is negative right up to $ \ p \ = \ 2 \ \ , $ it intersects the $ \ x-$axis for $ \ x \ < \ 0 \ $ only once right up to the transition from "upward-opening" parabola to the straight line $ \ -\frac72·x - \frac12 \ $ for $ \ p \ = \ \frac32 \ \ $ (with $ \ x-$intercept $ \ x \ = \ -\frac17 \ ) \ . $ The parabola "opens downward" with $ \ y-$intercept $ \ y \ < \ 0 \ \ $ up to $ \ p \ = \ 2 \ \ , $ so there are two $ x-$intercepts for $ \ \frac32 \ < \ p \ \le \ 2 \ , $ but only one is "to the left" of the $ \ y-$axis for $ \ p \ > \ 2 \ \ . $
Sorting this out, the situation for $ \ x \ \le \ 0 \ $ is that there is one intersection with $ \ y \ = \ 2 \ $ for $ \ p \ \le \ \frac32 \ \ , $ two intersections for $ \ \frac32 \ < \ p \ \le \ 2 \ \ , $ and one for $ \ p \ > \ 2 \ \ . $
We at last want to count the total number of intersections of the original curve $ \ y \ = \ (2p-3)x|x| \ + \ (3p-1)|x| \ + \ p \ \ $ with the line $ \ y \ = \ 2 \ \ . $ For the "special" values of $ \ p \ $ and the intervals "surrounding" them, we obtain
$ \mathbf{p \ < \ 1 \ \ :} \ \ 1 \ + \ 0 \ \ = \ \ 1 \ \ ; \ \ \mathbf{p \ = \ 1 \ \ :} \ \ 1 \ + \ 1 \ \ = \ \ 2 \ \ ; \ \ \mathbf{1 \ < \ p \ < \ \frac32 \ \ :} \ \ 1 \ + \ 2 \ \ = \ \ 3 \ \ ; $
$ \mathbf{p \ = \ \frac32 \ \ :} \ \ 1 \ + \ 1 \ \ = \ \ 2 \ \ ; \ \ \mathbf{\frac32 \ < \ p \ \le \ 2 \ \ :} \ \ 2 \ + \ 1 \ \ = \ \ 3 \ \ ; \ \ \mathbf{ p \ = \ 2 \ \ :} \ \ 1 \ + \ 1 \ \ = \ \ 2 \ \ $ (the "duplication" between the two cases is that both parabolas has their $ \ y-$intercept at $ \ y \ = \ 2 \ \ ) \ ; $
$ \mathbf{ p \ > \ 2 \ \ :} \ \ 1 \ + \ 0 \ \ = \ \ 1 \ \ . $
There doesn't seem to be a lot that can be done to avoid dealing with "cases", but it is a bit less maddening to look at what happens on "each side" of the $ \ y-$axis systematically as $ \ p \ $ is varied. To answer the question, the given equation has only one real solution for $ \ p \ < \ 1 \ $ or $ \ p \ > \ 2 \ \ . $