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Question Number 158410 by tebohlouis last updated on 03/Nov/21

Answered by MJS_new last updated on 03/Nov/21

$$2x+3⩾0\wedge (x-2)(x+1)>0\vee 2x+3⩽0\wedge (x-2)(x+1)<0$$$$$$$$2x+3⩾0\Rightarrow x⩾-\frac{3}{2}$$$$(x-2)(x+1)>0\Rightarrow x<-1\vee x>2$$$$\Rightarrow -\frac{3}{2}⩽x<-1\vee x>2$$$$$$$$2x+3⩽0\Rightarrow x⩽-\frac{3}{2}$$$$(x-2)(x+1)<0\Rightarrow -1<x<2$$$$\mathrm{no}\mathrm{solution}$$$$$$$$\Rightarrow x\in [-\frac{3}{2};-1)\cup (2;+\mathrm{\infty })$$

Answered by physicstutes last updated on 06/Nov/21

$${\mathrm{zero}}^{\prime }\mathrm{s}$$$$x-2=0\Rightarrow x=2$$$$x+1=0\Rightarrow x=-1$$$$2x+3=0\Rightarrow x=-\frac{3}{2}$$$$\overline{)\begin{array}{cccc}x<-\frac{3}{2}& -\frac{3}{2}<x<-1& -1<x<2& x>2\\ -& +& -& +\end{array}}$$$$S=\{x:-\frac{3}{2}⩽x<-1\cup x>2\}$$

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