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Roots of $ax^2 + bx + c = 0$ are real and positive. $a$, $b$ and $c$ are real.

Then $ax^2 + b|x| + c = 0$ has how many real roots?

My try:

I studied one method where we see how many signs are changing in equation. Then we are able to find real roots. But I forget the method name. So find difficult to solve.

Other methods are also appreciated. Thank you.

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    @Lovsovs I think that the first line is a condition for the problem – 2017-02-10
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    You know.... the answer would be the same if the roots weren't stated to be positive. – 2017-02-10
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    Seeing as you can't tell what signs a,b and c are that method is actually irrelevant to this problem. – 2017-02-10
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    @fleablood Let me see. With $x^2-3x+2$, the assumption that the roots be positive holds, and the modified equation $x^2-3|x|+2$ has ___four___ real roots, $\pm 1, \pm 2$. However with $x^2+x-2$, that assumption does not hold, and here the modified equation $x^2+|x|-2$ has ___two___ real roots, $\pm 1$. So how can you say "the answer would be the same if the roots weren't stated to be positive"? Of course when we go to $x^2+3x+2$, then $x^2+3|x|+2=0$ is impossible because for all real $x$ it is true that $x^2+3|x|+2 \ge 0+0+2 > 0$. – 2017-02-10
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    Are you denying that $j,k = -1, 1$ are roots then $-j, -k = 1,-1$ are also roots? You do have a point that I didn't distinguish uniqueness. But then the problem never stated that the equation did not have a double root (and, no, I do not consider the "s" at the end of roots to imply such.) The answer is still the same that if $j,k$ are roots then $j,k,-j,-k$ are also roots. That this may be $0, 1, 2 $ or $4$ is of minor relevance in my opinion. – 2017-02-10
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    @fleablood Please elaborate using $x^2+3x+2$. It has two distinct roots $j=-2$ and $k=-1$ and in fact $\{ +j, -j, +k, -k \}$ are four pair-wise distinct numbers. What do you say about the number of solutions to the equation $x^2+3|x|+2$ with absolute value (see my previous comment)? Did you note how the roots of the equation without absolute value were both negative, $j=-2<0$ and $k=-1<0$? Had $j$ and $k$ been positive as in the original question, this could not happen. – 2017-02-11

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Given that the roots of $ax^2+bx+c$ are both positive, and real, let the roots be $x=\alpha, \beta>0$ which follows from the condition. Note that the quadratic with the absolute value can be changed into $$ax^2+b|x|+c=0 \iff a|x|^2+b|x|+c=0$$ So we have$$|x|=\alpha, \beta$$ is a root. So there are four roots, $\alpha$, $-\alpha$, $\beta$, $-\beta$.

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    Can you please explain how you reach to conclusion that $|x| \to \alpha, \beta$ – 2017-02-10
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    @JohnSr We have that any $x$ that satisfies $ax^2+bx+c=0$ satisfies either $x= \alpha$ or $x=\beta$. In the second equation, since $t=|x|$ is an variable that satisfies $at^2+bt+c=0$, this, we have $t=|x|=\alpha$ or $\beta$. – 2017-02-10
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    @JohnSr Does thou understand? – 2017-02-10
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    how about graphical explanation. If $f(x)=ax^2+bx+c$ then the other one is $f(|x|)$ . To draw them you draw $f(x)$ and the other one overlaps on the right side (for positive x) and mirrors itself to the left. – 2017-02-10
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    @kazz I personally don't like graphical explanations. Visual proofs are harder for me to follow, and they seem handwavy. – 2017-02-10
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    how about graphical explanation. If $f(x)=ax^2+bx+c$ then the other one is $f(|x|)$ . To draw them you draw $f(x)$ and the other one overlaps on the right side (for positive x) and mirrors itself to the left. So if there is a root ot right side a new one shows up on the left. There doesn't has to be 4 root actually, there are TWICE as much $f(x)$ can be like $a(x-p)^2$ and will have only 1 (double) root. You may even argue if "roots are real" actually imply the they exist of not :P. Well long story short the proper answer is "Twice as much". – 2017-02-10
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    Well, then, as I said I don't care for graphical explanations. – 2017-02-10
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    hehe, ok proves by waving rapidly and erase the blackboard shortly after are not fun. – 2017-02-10
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    @Kazz $$?$$ $$?$$$$?$$ – 2017-02-10
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    How about this. Let $\alpha$ be a positive root of $f(x)$ therefore $a(-\alpha)^2+b|-\alpha|+c = a(\alpha)^2+b\alpha+c = f(\alpha) = 0$ which means that $-\alpha$ is in fact a root of the second function – 2017-02-10
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    @kazz Which is basically what I did. If you think you have another idea, you should add it as another answer? – 2017-02-10
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    ok. I was trying to address the question in the first comment. – 2017-02-10
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The solutions to this equation are the solutions to any of the equations : $ax^2+bx+c=0$ and $ax^2-bx+c=0$.

Hence the solutions to the following can be (at most $4$ values) : $$\frac { \pm b \pm \sqrt {b^2-4ac}}{2a} $$

Hope it helps.

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how about graphical explanation.

If $$f(x)=ax^2+bx+c$$ then the other one $$ax^2+b|x|+c = a|x|^2+b|x|+c = f(|x|)$$

To draw something like $f(|x|)$ you draw $f(x)$ on the right side (for positive x) and mirror it to the left. So if there is a root on right side a new one shows up on the left.

There doesn't has to be 4 root actually. $f(x)$f can be like $a(x−p)^2$ and will have only 1 (double) root.

You may even argue if "roots are real" actually imply the they exist of not.

Well long story short the proper answer is "Twice as much"

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It's a weird question but it boils down to: If $k > 0$ then $|k| = k$ and $|-k| = k$ and $(-k)^2 = k^2$ so if $k > 0$ is a solution to $ax^2 + bx + c$ then both $k$ and $-k$ are solutions to $ax^2 + b|x| + c$.

[Another way of putting it is $ax^2 + b|x| + c = a(-x)^2 + b|-x| + c$ so $k$ is a solution if and only if $-k$ is a solution. If $x > 0$ then $a^2 + bx + c = ax^2 + b|x| + c = a(-x)^2 + b|-x| + c$ then $k$ is a positive solution to $a^2 + bx + c$ if and only if $k, -k$ are a positive and negative solution to $ax^2 + b|x| + c$..]

So if $a^2 + bx + c$ has two positive real solutions; $k,j$ then $a^2 + b|x| + c$ will have the same two positive real solutions, $k,j$ as well as their negative counter parts $-k,-j$ as solutions.

... Note: $ax^2 + b|x| + c$ is NOT a polynomial. Hence the rule that there are at most two solutions does not apply. (Actually, $ax^2 + b|x| + c$ is a system of two polynomials: $ax^2 + bx + c$ or $ax^2 - bx + c$ and as each may have up to two solutions, the system will have up to four solutions [to one or the other]. Furthermore, if $k,j$ are the solutions to $ax^2 + bx + c$ then $-k,-j$ are the solutions to $ax^2 -bx +c$.)