Postive and negative real roots

Question

If ax^2 + bx + c has two real roots m and n such that m > 0 and n< 0 and |m| > |n| . Then how many real roots does ax^2 + b|x| + c have ?

To solve this question, I just took an example of x^2 -x - 42=0 with 2 roots 7 and -6.

Now for x>0 the equation remains the same as x^2 - x - 42 but with only one root 7 because x should be greater than 0.

For x<0, the example equation becomes x^2 + x -42 with root -7 (as x <0).

So I think the answer should be 2 real roots m and -m . But the answer is m and n.

Can someone please point out my mistake ?

Answer 1

The original equation is $ax^2+ bx+c=0$ with roots $m,n$, where $m>0$ and $n<0$ and $|m| > |n|$

The equation $ax^2+b|x| + c=0$ is equivalent to the two equations:

$$ax^2+ bx+c=0 , \, x \ge0\space\space\dots(1)\\ ax^2 -bx +c=0, \, x<0\space\space\dots(2)$$

1. The roots of $(1)$ are $m$ and $n$, as it is given in the question. But since the domain is restricted to positive real, $n$ is not a root of $(1)$ as $n <0$. So the only root is $m$.

2. The roots of $(2)$ are $-m$ and $-n$ as it is the equation formed after transformation $X \to-x$. But since the domain is restricted to negative reals, $-n$ cannot be a root as $-n > 0$. So the only root is $-m$.

So your answer that $m$ and $-m$ are the roots of the given equation is correct.

Answer 2

You are correct, and the given answer is wrong.

Your example is a counterexample to the given answer - see for example http://www.wolframalpha.com/input/?i=x%5E2-%7Cx%7C-42 for a quick and dirty verification of your calculation above.

We can show this more generally. The roots of $$ax^2+bx+c$$ are $$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ by the quadratic formula, and one root is negative and the other is positive, we must have $$m=\frac{-b + \sqrt{b^2-4ac}}{2a}>0\,\,;\,\, n=\frac{-b - \sqrt{b^2-4ac}}{2a}<0.$$ (By the way, if $|m|>|n|,$ then the above formulae tell us that $\frac{b}{2a}<0,$ although that fact will not be used in what follows.)

Using a similar argument to yours, one can figure out that there are exactly two roots of $$ax^2+b|x|+c:$$

for $x>0$, $m$ and for $x<0$, $\frac{b - \sqrt{b^2-4ac}}{2a} = -m.$

Answer 3

first try to draw all possible random graphs of the function $ax^2+bx+c$ having two roots let $f(x)=ax^2+bx+c$ then $ax^2+b|x|+c$ is $f(|x|)$ since $x^2=|x|^2$ so you invert part of graph which is -ve then you can see the roots of given $f(|x|)$ clearly