# Math Help - Could anyone please explain how to solve this absolute equations?

1. ## Could anyone please explain how to solve this absolute equations?

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!

2. Originally Posted by khushi
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 |

Thanks!
Case-I $x^2 + 3x \ge 0$ and $x^2 - 2x \ge 0$, then

$x^2 + 3x = x^2 - 2x$

$\Rightarrow x = 0.$

Case-II $x^2 + 3x \ge 0$ and $x^2 - 2x < 0$, then

$x^2 + 3x = -(x^2 - 2x)$

$\Rightarrow 2x^2 +x = 0.$

$\Rightarrow x = 0, \;\; \frac{-1}{2}$

Case-III $x^2 + 3x < 0$ and $x^2 - 2x \ge 0$, then

$-(x^2 + 3x) = x^2 - 2x$

$\Rightarrow 2x^2 +x = 0.$

$\Rightarrow x = 0, \;\; \frac{-1}{2}$

Case-IV $x^2 + 3x < 0$ and $x^2 - 2x < 0$, then

$-(x^2 + 3x) = -(x^2 - 2x)$

$\Rightarrow x = 0.$

3. Originally Posted by khushi
The problems ask to slove each inequality and express your answer using set notation or interval notation and then graph the solution set.

2. | -4x | + | -5 | <= 1

Thanks!
Case-I $-4x \ge 0$

$-4x + 5 \le 1$

$-4x \le 1-5$

$-4x \le -4$

$x \ge 1$

Case-II -4x < 0

$-(-4x) + 5 \le 1$

$4x \le -4$

$x \le -1$

You try the third question yourself.