Question

The solutions to the inequality (2x+3)^1/2 >lxl exist within which interval?

A. the interval -3/2<x<-1

B. the interval -x>3/2

C. the interval -3/2<x<3

Answer 1

$$2x+3\geq0=>x\geq-\dfrac{3}{2}$$

$$2x+3>x^2$$

$$x^2-2x-3<0$$

$$(x+1)(x-3)<0$$

$$-1<x<3, x\geq-\dfrac{3}{2}$$

The solutions are $-1<x<3.$

Interval $(-1,3)$ lies within the interval $-\dfrac{3}{2}<x<3.$