Finding $a$ for which $x^2a-2x+1=3\lvert x\rvert$ has exactly $3$ distinct real solutions. 
Find all real numbers a for which the equation $$x^2a-2x+1=3\lvert x\rvert$$ has exactly $3$ distinct real solutions in $x$.

I have tried the question very much but a big doubt is how a quadratic equation can have $3$ solutions.It is only possible when it is a identity in $x$. But I do not see here any identity. So please clear my doubt.
I know that $\lvert x\rvert$ can be solved as $+x$ and $-x$ taking both cases but nothing useful result was solved from here. I am unable to solve it further by this case.
So please answer the question.
Thanks
 A: $$ax^2-2x+1=3|x|\tag1$$
If $a\lt 0$, then the LHS of $(1)$ is a downward parabola and the RHS is V-shape, so $(1)$ cannot have three distinct real solutions.
In the following, $a\gt 0$.
If $x\ge 0$, then $$ax^2-5x+1=0\tag2$$ has at most two real solutions in $x\ge 0$, and it cannot have two real solutions $\alpha,\beta$ such that $\alpha\lt 0\lt \beta$ because the LHS of $(2)$ is positive when $x=0$.
If $x\lt 0$, then $$ax^2+x+1=0\tag3$$ has at most two real solutions in $x\lt 0$, and it cannot have two real solutions $\alpha,\beta$ such that $\alpha\lt 0\lt \beta$ because the LHS of $(2)$ is positive when $x=0$.
In order for $(1)$ to have three distinct real solutions, we have to have that either $(2)$ with $x\ge 0$ or $(3)$ with $x\lt 0$ has only one solution, which means that we have to have
$$(-5)^2-4a=0\qquad\text{or}\qquad 1^2-4a=0$$
i.e.
$$a=\frac{25}{4},\ \frac{1}{4}$$
For $a=\frac 14$, we have $x=-2,10\pm 4\sqrt 6$, so $a=\frac 14$ is sufficient.
For $a=\frac{25}{4}$, $(3)$ has no real solutions, so $a=\frac{25}{4}$ is not sufficient.
Therefore, $\color{red}{a=\frac 14}$ is the only answer.
A: Clearly $x=0$ is not a solution. Write $$\color{green}{f(x) = {2x+3|x|-1\over x^2}}$$
You are interested when a paralell $\color{red}{y=a}$ to $x$-axsis cuts the graph exactly $3$ times.  We see that hapens exactly when $a$ is local maximum for $x<0$ which is ${1\over 4}$, so the answer $a={1\over 4}$.

We can give from the graph complete analysis of the number of solution with respect to $a$:


*

*If $0<a<{1\over 4}$ it has $4$ solutions;

*If $a={1\over 4}$ it has $3$ solutions;

*If $a\leq 0$ or ${1\over 4}<a<{25\over 4}$ it has $2$ solutions;

*If $a={25\over 4}$ it has $1$ solutions;

*If ${25\over 4}<a$ it has no solutions.



A: Hint: For $$x\geq 0$$ we get
$$x^2a-5x+1=0$$ and the case $$a=0$$ can not be, so we get
$$x^2-\frac{5}{a}x+\frac{1}{a}=0$$
and we obtain
$$x_{1,2}=\frac{5}{2a}\pm\sqrt{\frac{25}{4a^2}-\frac{1}{a}}$$
Can you proceed?
A: It's obvious that $a>0$.
We need that $y=-3x$ will touch to parabola $y=ax^2-2x+1$ or
$y=3x$ will touch to parabola $y=ax^2-2x+1$.
Can you end it now?
The first case gives $a=\frac{1}{4}$ and it's valid.
The second case does not give solution for $a$.
A: 
One of the possibility with 3 distinct solutions when $a=0.25$
See here and slide the value of $a$ to examine various possibilities
A: Squaring and rearranging you have $$\left(ax^2-2x+1\right)^2-9x^2=0$$
So that $$(ax^2-5x+1)(ax^2+x+1)=0$$
And this has exactly three real solutions. Each quadratic factor has $0$ or $1$ or $2$ real roots (solutions to quadratic $=0$). The only way in which you can get exactly three solutions overall is for one of the factors to have a single real root, and this happens only if it is $\pm$ an exact square (or $a=0$, which can't give three solutions overall).
So the candidate solutions come from $\pm(px+q)^2=ax^2-5x+1$ and $\pm(px+q)^2=ax^2+x+1$, from which $q^2=1$ and then $2pq=-5, 1$ and $a=p^2$.
To solve directly square the second of these to obtain $4p^2q^2=4p^2=25, 1$ and $a=p^2$
Then these solutions need to be tested to see if they give two roots for the other factor, as required to bring the count up to $3$.
