Inequality rational polymonial

[xzi86]

Homework Statement

(3x+1)/(x+4)>=1

Homework Equations

The Attempt at a Solution

(3x+1)>=(x+4)
2x>=3
x>=3/2

But this is wrong?? Why?

 

[HallsofIvy]

If you multiply both sides of an inequality by a negative number, you reverse the direction of the inequality. But you don't know whether x+ 4 is positive or negative.

Instead break it into two parts. x+ 4 is positive, that is x+ 4> 0 for x> -4. If that is the case, then 3x+1> x+ 4 so that 2x> 3, x> 3/2. Because any value of x that is larger than 3/2 certainly is larger than -4, x>= 3/2 is part of the solution.

But if x< -4, then x+ 4 is negative. And then 3x+1< x+ 4 so that 2x< 3 so that x< 3/2. Now we argue the other way: any number that is less than -4 certainly is less than 3/2 so x< -4 is the other part of the solution. (x= -4 is not because the fraction is not defined if x= -4.)
You can check: if x= -5, then 3x+1= -15+1= -14, x+ 4= -1 so that (3x+1)/(x+4)= -14/-1= 14 which is certainly greater than 1.

That is, the solution set is all values of x less than -4 and all values of x greater than 3/2.

 

[xzi86]

so my method was good except i forgot to consider x is negative

 

[Mentallic]

so my method was good except i forgot to consider x is negative

Not x is negative, but x+4 is negative, or x+4<0, x<-4.

Another way to solve this problem without considering cases is to multiply by (x+4)² because any number squared is always positive, so there is no need to switch the inequality sign. You have to solve a quadratic and you'll get the same answer.

 

[BloodyFrozen]

Read this, it should help.


http://tutorial.math.lamar.edu/Classes/Alg/SolveRationalInequalities.aspx

In your case,

(3x+1)/(x+4)≥1


You want to get one side zero. Here is a small part...

(3x+1)/(x+4)≥1
(3x+1)/(x+4)-1≥0
Now, what can you do to make the LHS into one fraction?

If you read carefully, you should know.