Online Course Discussion Forum
Math Challenge II-A Problem 2.18
Hello! I was reviewing this lecture, and I came to problem 2.18. It asks for how many values is f(x) negative. I got that x<-3 or x>2 b/c of (x+3)(x-2)<0. How is there 4 values? I looked at the solutions and it said something about roots negative in between something. Can you help clarify? Thanks!
Tina
Hi Tina,
There's two ways to understand the final answer here.
I would recommend a hybrid algebraic and graphical approach. Factoring like you did, you get that the graph of $y=x^2+x-6$ has zeros at $x=-3$ and $x=2$. Since the graph has a positive $x^2$ coefficient, it opens upward, so the graph looks like this (click on the graph below to view on Desmos):
This is where the "between" comes from. From the graph we can see it will be negative for $x$ values with $-3 < x < 2$. This gives the $4$ values $x=-2, -1, 0, 1$.
For the purely algebraic approach, factoring gives $(x+3)(x-2) < 0$. Here, for the product of two numbers to be negative, we need one to be positive and the other to be negative. Hence,
- $x+3 < 0$ and $x-2 > 0$, implying that $x < -3$ and $x > 2$, but this is impossible.
- $x+3 > 0$ and $x - 2 < 0$, implying $x > -3$ and $x < 2$, so $-3 < x < 2$ as before.
For the purely algebraic approach, you do need to be careful with the "and", as we did in the two cases above. Hope this helps!
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