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Question Number 33836 by NECx last updated on 25/Apr/18

for what values of x if ((x(x−1))/(2x+3))>0

$${for}\:{what}\:{values}\:{of}\:{x}\:{if} \\ $$ $$\frac{{x}\left({x}−\mathrm{1}\right)}{\mathrm{2}{x}+\mathrm{3}}>\mathrm{0} \\ $$

Answered by Rasheed.Sindhi last updated on 26/Apr/18

((x(x−1))/(2x+3))>0 { ((x(x−1)>0 ∧ 2x+3>0...A)),((x(x−1)<0 ∧ 2x+3<0....B)) :} A : x>0 ∧ x−1>0 x>0 ∧ x>1 x>1 or x<0 ∧ x−1<0 x<0 ∧ x<1 x<0 Cont.

$$\frac{{x}\left({x}−\mathrm{1}\right)}{\mathrm{2}{x}+\mathrm{3}}>\mathrm{0\begin{cases}{{x}\left({x}−\mathrm{1}\right)>\mathrm{0}\:\wedge\:\mathrm{2}{x}+\mathrm{3}>\mathrm{0}...\mathrm{A}}\\{{x}\left({x}−\mathrm{1}\right)<\mathrm{0}\:\wedge\:\mathrm{2}{x}+\mathrm{3}<\mathrm{0}....\mathrm{B}}\end{cases}} \\ $$ $$\mathrm{A}\::\:\:{x}>\mathrm{0}\:\wedge\:{x}−\mathrm{1}>\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:{x}>\mathrm{0}\:\wedge\:{x}>\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:\:{x}>\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{or} \\ $$ $$\:\:\:\:\:\:{x}<\mathrm{0}\:\wedge\:{x}−\mathrm{1}<\mathrm{0} \\ $$ $$\:\:\:\:\:\:{x}<\mathrm{0}\:\wedge\:{x}<\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:{x}<\mathrm{0} \\ $$ $$\mathrm{Cont}. \\ $$

Commented byNECx 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

$${please}\:{I}\:{dont}\:{really}\:{understand} \\ $$ $${this}\:{style}.\:{Please}\:{shed}\:{more} \\ $$ $${light}\:{on}\:{the}\:{principle}\:{governing} \\ $$ $${this}\:{style}.{Thank}\:{you}\:{sir} \\ $$

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; ∞[

$$\mathrm{2}{x}+\mathrm{3}\neq\mathrm{0}\:\Rightarrow\:{x}\neq−\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\mathrm{1}.\:\mathrm{2}{x}+\mathrm{3}<\mathrm{0}\:\Rightarrow\:{x}<−\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $${x}\left({x}−\mathrm{1}\right)<\mathrm{0} \\ $$ $$\mathrm{1}.\mathrm{1}.\:{x}<\mathrm{0}\:\wedge\:{x}−\mathrm{1}>\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$ $$\mathrm{1}.\mathrm{2}.\:{x}>\mathrm{0}\:\mathrm{but}\:{x}<−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$ $$ \\ $$ $$\mathrm{2}.\:\mathrm{2}{x}+\mathrm{3}>\mathrm{0}\:\Rightarrow\:{x}>−\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $${x}\left({x}−\mathrm{1}\right)>\mathrm{0} \\ $$ $$\mathrm{2}.\mathrm{1}.\:{x}>\mathrm{0}\:\wedge\:{x}−\mathrm{1}>\mathrm{0}\:\Rightarrow\:{x}>\mathrm{1} \\ $$ $$\mathrm{2}.\mathrm{2}.\:{x}<\mathrm{0}\:\wedge\:{x}−\mathrm{1}<\mathrm{0}\:\Rightarrow\:{x}<\mathrm{0} \\ $$ $$ \\ $$ $$−\frac{\mathrm{3}}{\mathrm{2}}<{x}<\mathrm{0}\:\vee\:{x}>\mathrm{1} \\ $$ $$\left.{x}\in\right]−\frac{\mathrm{3}}{\mathrm{2}};\:\mathrm{0}\left[\:\cup\:\right]\mathrm{1};\:\infty\left[\right. \\ $$

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