# 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.

Thanks

• 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. – imranfat Jun 13 at 16:08
• what is desmos? – Aryan 24k Jun 13 at 16:10
• @imranfat what do you want to say from modulusx=sqrt of x^2. – Aryan 24k Jun 13 at 16:14
• 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 $x$. – imranfat Jun 13 at 16:45

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

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.

• how do you get that $alpha$ and $beta$ will not be of opposite signs – Aryan 24k Jun 14 at 9:59
• Why you do D=0. – Aryan 24k Jun 14 at 9:59
• @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. – mathlove Jun 14 at 10:30
• @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. – 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 $$0 it has $$4$$ solutions;
• If $$a={1\over 4}$$ it has $$3$$ solutions;
• If $$a\leq 0$$ or $${1\over 4} it has $$2$$ solutions;
• If $$a={25\over 4}$$ it has $$1$$ solutions;
• If $${25\over 4} it has no solutions.

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?

• 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 – Aryan 24k Jun 13 at 16:13
• @Aryan24k You will also have to consider $x\lt 0$ separately. – Vineet Jun 13 at 16:26
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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 $$a$$.

• why a is greater then 0 and why does y=1/3 touch the parabola. – Aryan 24k Jun 13 at 16:22
• @Aryan 24k Because we need three different roots. – Michael Rozenberg Jun 13 at 16:23
• @Vineet I don't agree with you. On your picture we have four real roots. – Michael Rozenberg Jun 13 at 16:28
• 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 – Vineet Jun 13 at 16:49
• Wow. I worked with $\frac{1}{3}|x|.$ Thank you! I'll fix my post. – 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 $$3$$.

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

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

• 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 – Aryan 24k Jun 13 at 16:20
• others have given beautiful answers. My answer is to show you how we can have this particular equation and $>2$ nos. of solutions. – Vineet Jun 13 at 16:22