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Question Number 151622 by otchereabdullai@gmail.com last updated on 22/Aug/21

Find the inequality with integer coefficient for the given solution set x:x < ((−2)/5) or >(2/5)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{with}\:\mathrm{integer}\: \\ $$ $$\mathrm{coefficient}\:\mathrm{for}\:\mathrm{the}\:\mathrm{given}\:\mathrm{solution}\:\mathrm{set} \\ $$ $$\mathrm{x}:\mathrm{x}\:<\:\frac{−\mathrm{2}}{\mathrm{5}}\:\mathrm{or}\:>\frac{\mathrm{2}}{\mathrm{5}} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Aug/21

x < ((−2)/5) or >(2/5) 5x<−2 or 5x>2 5x+2<0 or 5x−2>0 (5x+2)(5x−2)<0 25x^2 −4<0

$$\mathrm{x}\:<\:\frac{−\mathrm{2}}{\mathrm{5}}\:\mathrm{or}\:>\frac{\mathrm{2}}{\mathrm{5}} \\ $$ $$\mathrm{5x}<−\mathrm{2}\:\mathrm{or}\:\mathrm{5x}>\mathrm{2} \\ $$ $$\mathrm{5x}+\mathrm{2}<\mathrm{0}\:\mathrm{or}\:\mathrm{5x}−\mathrm{2}>\mathrm{0} \\ $$ $$\left(\mathrm{5x}+\mathrm{2}\right)\left(\mathrm{5x}−\mathrm{2}\right)<\mathrm{0} \\ $$ $$\mathrm{25x}^{\mathrm{2}} −\mathrm{4}<\mathrm{0} \\ $$

Commented bymr W last updated on 22/Aug/21

5x+2<0 or 5x−2>0 when 5x+2<0, then 5x−2<0 or when 5x−2>0, then 5x+2>0 in both cases: (5x+2)(5x−2)>0 not <0 i.e. 25x^2 −4>0 not <0

$$\mathrm{5x}+\mathrm{2}<\mathrm{0}\:\mathrm{or}\:\mathrm{5x}−\mathrm{2}>\mathrm{0} \\ $$ $${when}\:\mathrm{5}{x}+\mathrm{2}<\mathrm{0},\:{then}\:\mathrm{5}{x}−\mathrm{2}<\mathrm{0} \\ $$ $${or} \\ $$ $${when}\:\mathrm{5}{x}−\mathrm{2}>\mathrm{0},\:{then}\:\mathrm{5}{x}+\mathrm{2}>\mathrm{0} \\ $$ $${in}\:{both}\:{cases}: \\ $$ $$\left(\mathrm{5x}+\mathrm{2}\right)\left(\mathrm{5x}−\mathrm{2}\right)>\mathrm{0}\:{not}\:<\mathrm{0} \\ $$ $${i}.{e}.\:\mathrm{25}{x}^{\mathrm{2}} −\mathrm{4}>\mathrm{0}\:{not}\:<\mathrm{0} \\ $$

Commented bymr W last updated on 22/Aug/21

⇒ { ((x>−(2/5) & x>(2/5) ⇒ x>(2/5))),(( OR ⇒ or)),((x<−(2/5) & x<(2/5) ⇒ x<−(2/5) )) :} this is exactly x<−(2/5) or x>(2/5) which is requested. why you changed it to x>−(2/5) or x<(2/4) ?

$$\Rightarrow\begin{cases}{\mathrm{x}>−\frac{\mathrm{2}}{\mathrm{5}}\:\&\:{x}>\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\Rightarrow\:{x}>\frac{\mathrm{2}}{\mathrm{5}}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{or}}\\{{x}<−\frac{\mathrm{2}}{\mathrm{5}}\:\&\:{x}<\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\Rightarrow\:{x}<−\frac{\mathrm{2}}{\mathrm{5}}\:}\end{cases} \\ $$ $${this}\:{is}\:{exactly}\:{x}<−\frac{\mathrm{2}}{\mathrm{5}}\:{or}\:{x}>\frac{\mathrm{2}}{\mathrm{5}} \\ $$ $${which}\:{is}\:{requested}. \\ $$ $${why}\:{you}\:{changed}\:{it}\:{to} \\ $$ $${x}>−\frac{\mathrm{2}}{\mathrm{5}}\:{or}\:{x}<\frac{\mathrm{2}}{\mathrm{4}}\:? \\ $$

Commented byotchereabdullai@gmail.com last updated on 22/Aug/21

ok! i now see how to aproach such questions thanks a lot sir! Rasheed and Prof W

$$\mathrm{ok}!\:\mathrm{i}\:\mathrm{now}\:\mathrm{see}\:\mathrm{how}\:\mathrm{to}\:\mathrm{aproach}\:\mathrm{such}\: \\ $$ $$\mathrm{questions}\:\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{sir}!\:\mathrm{Rasheed}\: \\ $$ $$\mathrm{and}\:\mathrm{Prof}\:\mathrm{W} \\ $$

Commented byRasheed.Sindhi last updated on 22/Aug/21

mr W sir (5x+2)(5x−2)>0 ⇒ { ((5x+2>0 & 5x−2>0)),(( OR)),((5x+2<0 & 5x−2<0)) :} ⇒ { ((x>−(2/5) & x>(2/5))),(( OR)),((x<−(2/5) & x<(2/5))) :} ⇏x<^( ✓) −(2/5) or x<(2/5) Where′s error sir?

$${mr}\:\mathrm{W}\:\boldsymbol{{sir}} \\ $$ $$\left(\mathrm{5x}+\mathrm{2}\right)\left(\mathrm{5x}−\mathrm{2}\right)>\mathrm{0} \\ $$ $$\Rightarrow\begin{cases}{\mathrm{5x}+\mathrm{2}>\mathrm{0}\:\&\:\mathrm{5}{x}−\mathrm{2}>\mathrm{0}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR}}\\{\mathrm{5}{x}+\mathrm{2}<\mathrm{0}\:\&\:\mathrm{5}{x}−\mathrm{2}<\mathrm{0}}\end{cases} \\ $$ $$\Rightarrow\begin{cases}{\mathrm{x}>−\frac{\mathrm{2}}{\mathrm{5}}\:\&\:{x}>\frac{\mathrm{2}}{\mathrm{5}}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR}}\\{{x}<−\frac{\mathrm{2}}{\mathrm{5}}\:\&\:{x}<\frac{\mathrm{2}}{\mathrm{5}}}\end{cases} \\ $$ $$\nRightarrow{x}\overset{\:\:\checkmark} {<}−\frac{\mathrm{2}}{\mathrm{5}}\:{or}\:{x}<\frac{\mathrm{2}}{\mathrm{5}} \\ $$ $${Where}'{s}\:{error}\:{sir}? \\ $$

Commented byRasheed.Sindhi last updated on 22/Aug/21

T_H_A_N_K_S A L_(O_T ) S_(I_R_! ) I corrected my mistake.

$$\underset{\underset{\underset{\underset{\underset{\boldsymbol{{S}}} {\boldsymbol{{K}}}} {\boldsymbol{{N}}}} {\boldsymbol{\mathrm{A}}}} {\mathrm{H}}} {\boldsymbol{{T}}}\:\:\boldsymbol{{A}}\:\:\underset{\underset{\boldsymbol{{T}}} {\boldsymbol{{O}}}} {\boldsymbol{{L}}}\:\:\underset{\underset{\underset{!} {\boldsymbol{{R}}}} {\boldsymbol{{I}}}} {\boldsymbol{{S}}}\:\:{I}\:{corrected}\:{my}\:{mistake}. \\ $$

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