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?
By looking at your working I see you mean to solve $\displaystyle \frac{x-4}{x^2+3x} \geq 0$. You should use brackets to be more clear in future please...
It should be $\displaystyle x \in \left(-3,0\right) \cup \left[\,4, \infty \right)$
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.
Hello, makaveli89!
Solve: .$\displaystyle \frac{x-4}{x^2 + 3x}\:\geq \:0$
$\displaystyle x-4 \:\geq\:0 \quad\Rightarrow\quad x \:\geq\:4$
$\displaystyle x(x+3)\:\geq\:0 \quad\Rightarrow\quad x\:\geq\:0\:\text{ and }\:x\:\geq\:-3$
Thus: .$\displaystyle (-3,0]\:\text{ and }\:(4, \infty)$
Is this correct? . . . . no
You have: .$\displaystyle \begin{array}{c}x \geq 4 \\ x \geq 0 \\ x \geq \text{-}3 \end{array}\quad \hdots $ which means: .$\displaystyle x \geq 4\:\text{ or }\:[4,\infty)$ . . . and that's all.
That fraction is positive if both the numerator and denominator are negtive.
Numerator is negative: .$\displaystyle x-4 \:\leq \:0 \quad\Rightarrow\quad x \:\leq\:4$
Denominator is negative: .$\displaystyle x(x+3) \:< \:0$
There are two cases to consider:
$\displaystyle [1]\;\;\begin{array}{ccccccc}x > 0 & \Rightarrow & x\text{ positive} \\ x+3 < 0 & \Rightarrow & x < -3 \end{array}\;\;\hdots\;\text{ impossible}$
$\displaystyle [2]\;\;\begin{array}{cccccc}x < 0 & \Rightarrow & x\text{ negative} \\
x + 3 > 0 & \Rightarrow& x > \text{-}3 \end{array}$
We have: .$\displaystyle \begin{array}{c}x \:\leq \:4 \\ x \:<\:0 \\ x \:> \:\text{-}3 \end{array}\quad\hdots$ which means: .$\displaystyle -3 < x < 0 \:\text{ or }\;(-3,0)$
The answer is: .$\displaystyle (\text{-}3,0) \,\cup\, [4,\infty)$