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.
On
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.
On
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"
On
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 $ax^2 -bx +c$.)
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, $\alpha$, $-\alpha$, $\beta$, $-\beta$.