Question Number 33836 by NECx last updated on 25/Apr/18
$${for}\:{what}\:{values}\:{of}\:{x}\:{if} \\ $$$$\frac{{x}\left({x}−\mathrm{1}\right)}{\mathrm{2}{x}+\mathrm{3}}>\mathrm{0} \\ $$
Answered by Rasheed.Sindhi last updated on 26/Apr/18
$$\frac{{x}\left({x}−\mathrm{1}\right)}{\mathrm{2}{x}+\mathrm{3}}>\mathrm{0\begin{cases}{{x}\left({x}−\mathrm{1}\right)>\mathrm{0}\:\wedge\:\mathrm{2}{x}+\mathrm{3}>\mathrm{0}…\mathrm{A}}\\{{x}\left({x}−\mathrm{1}\right)<\mathrm{0}\:\wedge\:\mathrm{2}{x}+\mathrm{3}<\mathrm{0}….\mathrm{B}}\end{cases}} \\ $$$$\mathrm{A}\::\:\:{x}>\mathrm{0}\:\wedge\:{x}−\mathrm{1}>\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{x}>\mathrm{0}\:\wedge\:{x}>\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:{x}>\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{or} \\ $$$$\:\:\:\:\:\:{x}<\mathrm{0}\:\wedge\:{x}−\mathrm{1}<\mathrm{0} \\ $$$$\:\:\:\:\:\:{x}<\mathrm{0}\:\wedge\:{x}<\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{x}<\mathrm{0} \\ $$$$\mathrm{Cont}. \\ $$
Commented by NECx last updated on 26/Apr/18
$${please}\:{I}\:{dont}\:{really}\:{understand} \\ $$$${this}\:{style}.\:{Please}\:{shed}\:{more} \\ $$$${light}\:{on}\:{the}\:{principle}\:{governing} \\ $$$${this}\:{style}.{Thank}\:{you}\:{sir} \\ $$
Answered by MJS last updated on 26/Apr/18
![2x+3≠0 ⇒ x≠−(3/2) 1. 2x+3<0 ⇒ x<−(3/2) x(x−1)<0 1.1. x<0 ∧ x−1>0 ⇒ no solution 1.2. x>0 but x<−(3/2) ⇒ no solution 2. 2x+3>0 ⇒ x>−(3/2) x(x−1)>0 2.1. x>0 ∧ x−1>0 ⇒ x>1 2.2. x<0 ∧ x−1<0 ⇒ x<0 −(3/2)<x<0 ∨ x>1 x∈]−(3/2); 0[ ∪ ]1; ∞[](https://www.tinkutara.com/question/Q33854.png)
$$\mathrm{2}{x}+\mathrm{3}\neq\mathrm{0}\:\Rightarrow\:{x}\neq−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{1}.\:\mathrm{2}{x}+\mathrm{3}<\mathrm{0}\:\Rightarrow\:{x}<−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}\left({x}−\mathrm{1}\right)<\mathrm{0} \\ $$$$\mathrm{1}.\mathrm{1}.\:{x}<\mathrm{0}\:\wedge\:{x}−\mathrm{1}>\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{1}.\mathrm{2}.\:{x}>\mathrm{0}\:\mathrm{but}\:{x}<−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$ \\ $$$$\mathrm{2}.\:\mathrm{2}{x}+\mathrm{3}>\mathrm{0}\:\Rightarrow\:{x}>−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}\left({x}−\mathrm{1}\right)>\mathrm{0} \\ $$$$\mathrm{2}.\mathrm{1}.\:{x}>\mathrm{0}\:\wedge\:{x}−\mathrm{1}>\mathrm{0}\:\Rightarrow\:{x}>\mathrm{1} \\ $$$$\mathrm{2}.\mathrm{2}.\:{x}<\mathrm{0}\:\wedge\:{x}−\mathrm{1}<\mathrm{0}\:\Rightarrow\:{x}<\mathrm{0} \\ $$$$ \\ $$$$−\frac{\mathrm{3}}{\mathrm{2}}<{x}<\mathrm{0}\:\vee\:{x}>\mathrm{1} \\ $$$$\left.{x}\in\right]−\frac{\mathrm{3}}{\mathrm{2}};\:\mathrm{0}\left[\:\cup\:\right]\mathrm{1};\:\infty\left[\right. \\ $$