# How do you solve |5x+2|>=|3x-4|?

Jun 12, 2017

Solution : $x \le - 3 \mathmr{and} x \ge \frac{1}{4}$ OR $\left(- \infty , - 3\right] \cup \left[\frac{1}{4} , \infty\right)$

#### Explanation:

$| 5 x + 2 | \ge | 3 x - 4 |$ squaring both sides we get
${\left(5 x + 2\right)}^{2} \ge {\left(3 x - 4\right)}^{2}$ or
$25 {x}^{2} + 20 x + 4 \ge 9 {x}^{2} - 24 x + 16$ or
$25 {x}^{2} + 20 x + 4 - 9 {x}^{2} + 24 x - 16 \ge 0$ or
$16 {x}^{2} + 44 x - 12 \ge 0$ or
$4 {x}^{2} + 11 x - 3 \ge 0$ or
$\left(4 x - 1\right) \left(x + 3\right) . \ge 0$. Critical points are $x = - 3 \mathmr{and} x = \frac{1}{4}$
Sign Change:
For x=-3 or x=1/4; (4x-1)(x+3)=0
For x<-3 ; (4x-1)(x+3)= (-*-)= (+) ; (4x-1)(x+3)>0

For -3 < x < 1/4 ; (4x-1)(x+3)= (-*+)= (-) ; (4x-1)(x+3)<0

For x>1/4 ; (4x-1)(x+3)= (+*+)= (+) ; (4x-1)(x+3)>0
So solution : $x \le - 3 \mathmr{and} x \ge \frac{1}{4}$ or $\left(- \infty , - 3\right] \cup \left[\frac{1}{4} , \infty\right)$
The graph also confirms the findings.
graph{4x^2+11x-3 [-40, 40, -20, 20]}