3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $a$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @Vineet I don't agree with you. On your picture we have four real roots. $\endgroup$ – Michael Rozenberg Jun 13 at 16:28
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ Wow. I worked with $\frac{1}{3}|x|.$ Thank you! I'll fix my post. $\endgroup$ – Michael Rozenberg Jun 13 at 16:55
1
$\begingroup$

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 $3$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.