Solving an absolute value.

  • Thread starter AznBoi
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  • #1
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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
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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
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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
uart
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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
symbolipoint
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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
HallsofIvy
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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
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:smile:
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:
 

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