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Find-the-inequality-with-integer-coefficient-for-the-given-solution-set-x-x-lt-2-5-or-gt-2-5-




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 by mr 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 by mr 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 by otchereabdullai@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 by Rasheed.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 by Rasheed.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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