task is x-4 / x^2 + 3x >= 0 solve inequality

x-4 >= 0

x >= 4

x (x+3) , x=0 & x>= -3

thus, (-3,0] and (4, infinity)

is this correct?

Printable View

- April 11th 2009, 04:55 PMmakaveli89Solve inequality
task is x-4 / x^2 + 3x >= 0 solve inequality

x-4 >= 0

x >= 4

x (x+3) , x=0 & x>= -3

thus, (-3,0] and (4, infinity)

is this correct? - April 11th 2009, 05:27 PMnzmathman
By looking at your working I see you mean to solve . You should use brackets to be more clear in future please...

It should be

0 is not included in the first interval as that would cause division by zero. 4 is included in the second interval as we are solving for greater than or equal to 0.

EDIT: sorry fixed a mistake. - April 11th 2009, 05:58 PMSoroban
Hello, makaveli89!

Quote:

Solve: .

Thus: .

Is this correct? . . . . no

You have: . which means: . . . .*and that's all.*

That fraction is positive if both the numerator and denominator are negtive.

Numerator is negative: .

Denominator is negative: .

There are two cases to consider:

We have: . which means: .

The answer is: .