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When dealing with problems containing quadratic equations, I've never come across one that deals with a number of roots that is bigger than 2.

I've tried breaking up the function into two parts.

$x^2+6x+4-m=0$ for $x^2+5x+4\ge0$
and
$-x^2-4x-4-m=0$ for $x^2+5x+4<0$

Now imagining the graph of the function $\lvert x^2+5x+4\rvert$, I came to the idea that the equation in the problem should have more than 2 roots if at least one of the roots of the absolute part is smaller than 0 (under the x axis, thus 2+ roots).

The problem is, doing those two inequalities ($f(x_1)<0$ and $f(x_2)<0$), I get that m is greater than $-1$ or $-4$. I'm sure I'm on the right track because a few of the possible intervals contain these two numbers but I'm not sure where I made a mistake.

Edit:

The possible solutions are
A: $[1,4]$
B: $[-1,0]$
C: $[0,4]$
D: $[-4,-1]$

John Doe
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2 Answers2

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I think that drawing the graph of $y=x+|x^2+5x+4|$ will help.

This is because the number of real roots of $|x^2+5x+4|+x-m=0$ equals the number of the intersection points between $y=m$ and $y=x+|x^2+5x+4|$.

Since $y=m$ is a horizontal line, the graph should tell you the answer. As a result, the answer will be B. I hope this helps.

mathlove
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  • Would you mind explaining the second sentence, I'm not sure I follow? – John Doe Apr 20 '15 at 20:09
  • Understood, thank you :) Say you got this problem and weren't able to graph the function, is there any way to get this result without looking at it? I'm asking because in the setting I will be solving problems like this I most likely won't have access to a graph of a function like this (and I wouldn't know to graph it by hand). – John Doe Apr 20 '15 at 20:18
  • @JohnDoe: An easy example : $x^2=4$ can be interpreted as the $x$ coordinate of the intersection points between the graph of $y=x^2$ and the graph of $y=4$. Sorry:) I deleted my comment to improve it. – mathlove Apr 20 '15 at 20:18
  • @JohnDoe: Well, I think you can get the graphs. Just draw $y=x^2+6x+4$ for $x\le -4$ or $x\ge -1$ and $y=-x^2-4x-4$ for $-4\le x\le -1$. Don't you know how to draw parabolas? – mathlove Apr 20 '15 at 20:22
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    Welp! I do, for some reason that -x outside the absolute value was scaring me, never had to graph something like that, but you are right, it's not far more complex than a simple parabola.
    Thanks!     – John Doe Apr 20 '15 at 20:24
  • @JohnDoe: You are welcome! Yes, you can do it. Very glad if it helps. – mathlove Apr 20 '15 at 20:25
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Let $f(x) = \vert x^2+5x+4 \vert + x$. We then have \begin{align} f(x) & = \begin{cases} x^2+6x+4 & x \notin (-4,-1)\\ -(x+2)^2 & x \in (-4,-1) \end{cases} \end{align}

enter image description here

Hence, $g(x) = f(x) - m$ shifts the graph down by $m$. We see that $f(x)$ has $3$ roots.

  1. Shifting down any bit will reduce the number of roots to $2$. Hence, $m > 0$ results in two roots for $g(x)$.
  2. Also, note that shifting up by bringing the sharp peak at $x=-1$ above the $X$ axis but the sharp peak at $x=-4$ remaining below the $X$ axis also produces $2$ roots. The value of $f(x)$ at $x=-1$ is -1, while the value at $x=-4$ is $-4$. Hence, for any $m \in (-4,-1)$ we again have two roots.

Hence, to summarize:

  1. $m>0$ produces $2$ roots.

  2. $m=0$ produces $3$ roots.

  3. $m \in (-1,0)$ produces $4$ roots.

  4. $m=-1$ produces $3$ roots.

  5. $m \in (-4,-1)$ produces $2$ roots.

  6. $m=-4$ produces $1$ root.

  7. $m<-4$ produces no root.

Adhvaitha
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  • Thank you. The question however, is asking when the function will have *more* than two roots so the answer would be $[-1,0]$.

    Had I not been able to get the graph of this function, how would I go around to getting this result?

     – John Doe Apr 20 '15 at 20:12