# Solving an absolute value.

I'm confused by why my work is wrong:

$$3x^{2}+12x>0$$
$$3x(x+4)>0$$

3x>0 , x+4>0

x>0 , x>-4

However, the correct answers are: x>0 and x<-4??? Why is the sign reversed on -4? I thought you were only suppose to reverse it if you multiply or divide the equation by a negative number?

## Answers and Replies

x+4 is less than 0 when x is less than -4.
So you have to check all cases.

So, if x+4>0, then 3x>0 which gives x>0. But also x>-4. So the intersection of these is x>0.
Now try it for x+4<0.

Then do the same for 3x.

I think you can see there's no real x that satisfies x>0 and x<-4.

I'd say the solution is x>0 or x<-4.

It can be arrived at by breaking the problem into two cases:

1. 3x>0 and (x+4)>0. Clearly, in this case x>0 and x>-4 which implies that x must be greater than 0.

2. 3x<0 and (x+4)<0. Clearly, in this case x<0 and x<-4 which implies that x must be less than -4.

Conclusion: The inequality holds if either 1. or 2. are satisfied. Therefore, x>0 or x<-4.

Incidently, it's not clear to me why you refer to your problem as an absolute value problem.
I'd say you're just trying to solve an inequality (by an exhaustion of cases treatment).

uart
Science Advisor
One of the easiest ways to understand that type of problem is simply to visualize the parabola of the original inequality. Equation $$3x^2 + 12x$$ is a positive (or upward) parabola right! So just think about where, in relation to its zeros, that such a parabola sits above the x axis.

It's pretty easy if you think about it like that isn't it.

symbolipoint
Homework Helper
Education Advisor
Gold Member
The factoring was good. The best approach is to look for critical points (the values of x for which the expression becomes 0). Check a value for x in each interval and determine the truthfulness of the relation.

You can easily keep to one dimension, a single number line graph, for this kind of exercise.

HallsofIvy
Science Advisor
Homework Helper
If AB= 0 then A= 0 or B= 0

If AB> 0 it does NOT follow that A> 0 or B> 0. It might happen that A< 0 and B< 0.

A good way to solve general inequalities is to solve the associated equation first. For your example, 3x2+ 12x> 0, the associated equality is 3x2+ 12x= 3x(x+ 4)= 0 has roots x= 0, x= -4. That divides the real line into 3 intervals: $(-\infty, -4)$, (-4, 0), and $(0, \infty)$. If x< -4, then both x and x+4 are negative: their product is positive. If -4< x< 0, then x is still negative but x+ 4 is positive: their product is negative. If x> 0, then both x and x+ 4 are positive: their product is positive. 3x2+ 12x> 0 is satisfied for x< -4 or x> 0.

Why, by the way, was this titled "solving an absolute value"? If AB= 0 then A= 0 or B= 0

If AB> 0 it does NOT follow that A> 0 or B> 0. It might happen that A< 0 and B< 0.

A good way to solve general inequalities is to solve the associated equation first. For your example, 3x2+ 12x> 0, the associated equality is 3x2+ 12x= 3x(x+ 4)= 0 has roots x= 0, x= -4. That divides the real line into 3 intervals: $(-\infty, -4)$, (-4, 0), and $(0, \infty)$. If x< -4, then both x and x+4 are negative: their product is positive. If -4< x< 0, then x is still negative but x+ 4 is positive: their product is negative. If x> 0, then both x and x+ 4 are positive: their product is positive. 3x2+ 12x> 0 is satisfied for x< -4 or x> 0.

Why, by the way, was this titled "solving an absolute value"?

Thanks for all the replies. My mistake, I titled it that because it had a > sign xP I dunno, many absolute value problems have the signs lol. I think I need to go back and look at this. I've forgotten how to do this and I know that I've learned it last year. It makes sense to me though. Thanks for your help! 