# Inequalities

• Jul 1st 2009, 05:15 AM
cloud5
Inequalities
Can anybody help me to solve this?

1. $\frac{6}{\mid x \mid + 1} < \mid x \mid$

2. $\frac{6}{\mid x \mid} < \mid 2x-1 \mid$

3. $\mid x \mid < \mid x^2-7x+6 \mid$

1. $x<-2, x>2$

2. $\{x:x<-\frac{3}{2}\} \cup \{x:x>2\}$

3. $x<4-\sqrt{10}, 3-\sqrt{3}4+\sqrt{10}$
• Jul 1st 2009, 05:35 AM
Jester
Quote:

Originally Posted by cloud5
Can anybody help me to solve this?

1. $\frac{6}{\mid x \mid + 1} < \mid x \mid$

2. $\frac{6}{\mid x \mid} < \mid 2x-1 \mid$

1. $x<-2, x>2$
2. $\{x:x<-\frac{3}{2}\} \cup \{x:x>2\}$
1. $\frac{6}{\mid x \mid + 1} < \mid x \mid$ or $0 < \mid x \mid^2 + \mid x \mid - \,6\;\;\;\Rightarrow\;\;\; 0 < \left( \mid x \mid + 3\right)\left( \mid x \mid -2\right)$ which only happens if $\mid x \mid < -3$ (never) or $\mid x \mid > 2$, your answer.
2. $\frac{6}{\mid x \mid} < \mid 2x-1 \mid$ or $\mid 2x-1 \mid \mid x \mid > 6$ or $\mid 2x^2 - x \mid > 6$. Solutions of this are $2x^2-x > 6$ or $2x^2-x <-6$ (never). For the first $2x^2-x > 6$ or $(2x+3)(x-2) > 0$ which gives $x < - \frac{3}{2}$ and $x > 2$.