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
Best Answer
$$ax^2-2x+1=3|x|\tag1$$
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.