The problems ask to slove each inequality and express your answer using set notation or interval notation and then graph the solution set.
1. | x^2 + 3x | = | x^2- 2x |
2. | -4x | + | -5 | <= 1
3. 5 + | x - 1 | > 1/2
Thanks!
The problems ask to slove each inequality and express your answer using set notation or interval notation and then graph the solution set.
1. | x^2 + 3x | = | x^2- 2x |
2. | -4x | + | -5 | <= 1
3. 5 + | x - 1 | > 1/2
Thanks!
Case-I $\displaystyle x^2 + 3x \ge 0$ and $\displaystyle x^2 - 2x \ge 0$, then
$\displaystyle x^2 + 3x = x^2 - 2x$
$\displaystyle \Rightarrow x = 0.$
Case-II $\displaystyle x^2 + 3x \ge 0$ and $\displaystyle x^2 - 2x < 0$, then
$\displaystyle x^2 + 3x = -(x^2 - 2x)$
$\displaystyle \Rightarrow 2x^2 +x = 0.$
$\displaystyle \Rightarrow x = 0, \;\; \frac{-1}{2}$
Case-III $\displaystyle x^2 + 3x < 0$ and $\displaystyle x^2 - 2x \ge 0$, then
$\displaystyle -(x^2 + 3x) = x^2 - 2x$
$\displaystyle \Rightarrow 2x^2 +x = 0.$
$\displaystyle \Rightarrow x = 0, \;\; \frac{-1}{2}$
Case-IV $\displaystyle x^2 + 3x < 0$ and $\displaystyle x^2 - 2x < 0$, then
$\displaystyle -(x^2 + 3x) = -(x^2 - 2x)$
$\displaystyle \Rightarrow x = 0.$
Case-I $\displaystyle -4x \ge 0$
$\displaystyle -4x + 5 \le 1$
$\displaystyle -4x \le 1-5$
$\displaystyle -4x \le -4$
$\displaystyle x \ge 1$
Case-II -4x < 0
$\displaystyle -(-4x) + 5 \le 1$
$\displaystyle 4x \le -4$
$\displaystyle x \le -1$
You try the third question yourself.