Why is the sign reversed on -4 in solving 3x^{2}+12x>0?

In summary, the conversation discusses solving the inequality 3x^2 + 12x > 0 using factoring and breaking it into two cases. The solution is x > 0 or x < -4. It is mentioned that the associated equality is 3x^2 + 12x = 0, which has roots at x = 0 and x = -4. The real line is divided into three intervals and it is explained how to determine the truthfulness of the inequality for each interval. It is also mentioned that the title of the conversation is a mistake, as the problem is not an absolute value problem.
  • #1
AznBoi
471
0
I'm confused by why my work is wrong:

[tex]3x^{2}+12x>0[/tex]
[tex]3x(x+4)>0[/tex]

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?
 
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  • #2
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.
 
  • #3
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).
 
  • #4
One of the easiest ways to understand that type of problem is simply to visualize the parabola of the original inequality. Equation [tex]3x^2 + 12x[/tex] 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.
 
  • #5
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.
 
  • #6
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: [itex](-\infty, -4)[/itex], (-4, 0), and [itex](0, \infty)[/itex]. 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"?
 
  • #7
:smile:
HallsofIvy said:
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: [itex](-\infty, -4)[/itex], (-4, 0), and [itex](0, \infty)[/itex]. 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! :smile:
 

1. What is an absolute value?

An absolute value is a mathematical concept that represents the distance of a number from 0 on a number line. It is always a non-negative value, meaning it is either 0 or a positive number.

2. How do I solve an absolute value equation?

To solve an absolute value equation, you first isolate the absolute value expression on one side of the equation. Then, you create two separate equations, one with the absolute value expression set equal to a positive value and one with it set equal to a negative value. Solve both equations separately to find the two possible solutions.

3. Can absolute value equations have more than two solutions?

No, absolute value equations can only have two solutions at most. This is because the absolute value of a number can only have two possible values: the positive value or the negative value. Any other value would not be considered an absolute value.

4. How do I graph an absolute value function?

To graph an absolute value function, first identify the vertex of the graph, which is the point where the absolute value expression equals 0. Then, plot points on either side of the vertex by substituting different values for the variable in the absolute value expression. Finally, connect the points with a straight line to create the V-shaped graph.

5. Can absolute value equations have imaginary solutions?

No, absolute value equations can only have real number solutions. This is because the absolute value of a number is always a non-negative value, and imaginary numbers have a negative square root. Therefore, they cannot be solutions to absolute value equations.

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