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.


  • $\begingroup$ Actually, your equation can have even 4 solutions. Use Desmos to see how (ex: $a=0.1$). And in fact, since $|x|=\sqrt{x^2}$, through squaring, your equation becomes a quartic. So this is an interesting question indeed. $\endgroup$ – imranfat Jun 13 at 16:08
  • $\begingroup$ what is desmos? $\endgroup$ – Aryan 24k Jun 13 at 16:10
  • $\begingroup$ @imranfat what do you want to say from modulusx=sqrt of x^2. $\endgroup$ – Aryan 24k Jun 13 at 16:14
  • $\begingroup$ Part 1, check out www.desmos.com, it is a graphing tool, very nice. Part 2, $|x|=\sqrt{x^2}$ is a way to write the absolute value of $x$. $\endgroup$ – imranfat Jun 13 at 16:45


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.

  • $\begingroup$ how do you get that $alpha$ and $beta$ will not be of opposite signs $\endgroup$ – Aryan 24k Jun 14 at 9:59
  • $\begingroup$ Why you do D=0. $\endgroup$ – Aryan 24k Jun 14 at 9:59
  • $\begingroup$ @Aryan 24k : I should have mentioned that $a$ has to be positive. Now, let $f(x)$ be the LHS of $(2)$. Then, we get $f(0)=1$ which is positive. This means that $(2)$ cannot have two real solutions $\alpha,\beta$ such that $\alpha\lt 0\lt\beta$. (Considering the upward parabola $y=f(x)$ should help) This can be said for $(3)$ as well. $\endgroup$ – mathlove Jun 14 at 10:30
  • $\begingroup$ @Aryan24k: $(2)$ has at most two real solutions in $x\ge 0$, and $(3)$ has at most two real solutions in $x\lt 0$. In order for $(1)$ to have three distinct real solutions, from the comment above, it is necessary that either $D=0$ for $(2)$ or $D=0$ for $(3)$. I hope this helps. $\endgroup$ – mathlove Jun 14 at 15:18

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.

enter image description here


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?

  • $\begingroup$ but if I put a value of any a belonging to the correct answer then also x will have 2 values only as the polynomial is quadratic and it cannot have more then 2 values.So please elaborate $\endgroup$ – Aryan 24k Jun 13 at 16:13
  • 1
    $\begingroup$ @Aryan24k You will also have to consider $x\lt 0$ separately. $\endgroup$ – Vineet Jun 13 at 16:26
  • $\begingroup$ Thank you my dear friend from Leipzig. $\endgroup$ – Aqua Jun 27 at 12:41
  • $\begingroup$ Thank you for the flowers! $\endgroup$ – Dr. Sonnhard Graubner Jun 27 at 12:46

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 $a$.

  • $\begingroup$ why a is greater then 0 and why does y=1/3 touch the parabola. $\endgroup$ – Aryan 24k Jun 13 at 16:22
  • $\begingroup$ @Aryan 24k Because we need three different roots. $\endgroup$ – Michael Rozenberg Jun 13 at 16:23
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    $\begingroup$ @Vineet I don't agree with you. On your picture we have four real roots. $\endgroup$ – Michael Rozenberg Jun 13 at 16:28
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    $\begingroup$ but @MichaelRozenberg we still don't get 3 points of intersection if we use a = 49/36 & 25/36. Am I missing something? check the graph here desmos.com/calculator/4jt4bgydyu $\endgroup$ – Vineet Jun 13 at 16:49
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    $\begingroup$ Wow. I worked with $\frac{1}{3}|x|.$ Thank you! I'll fix my post. $\endgroup$ – Michael Rozenberg Jun 13 at 16:55

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 $3$.


enter image description here

One of the possibility with 3 distinct solutions when $a=0.25$

See here and slide the value of $a$ to examine various possibilities

  • $\begingroup$ Actually this question came in my exam and I can't use a graphics calculator so please can you tell how you get the value of a $\endgroup$ – Aryan 24k Jun 13 at 16:20
  • $\begingroup$ others have given beautiful answers. My answer is to show you how we can have this particular equation and $>2$ nos. of solutions. $\endgroup$ – Vineet Jun 13 at 16:22

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