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Question Number 57389 by rahul 19 last updated on 03/Apr/19
$${If}\:{a}\epsilon{R}\:{and}\:{the}\:{equation}\:: \\ $$$$−\mathrm{3}\left\{{x}\right\}^{\mathrm{2}} +\mathrm{2}\left\{{x}\right\}+{a}^{\mathrm{2}} =\mathrm{0}\:{has}\:{no}\:{integral} \\ $$$${solution},\:{then}\:{all}\:{possible}\:{value}\:{of}\:{a} \\ $$$${lie}\:{in}\:{the}\:{interval}\:: \\ $$$$\left({a}\right)\left(−\mathrm{1},\mathrm{0}\right)\mathrm{U}\left(\mathrm{0},\mathrm{1}\right)\:\:\:\left({b}\right)\left(\mathrm{1},\mathrm{2}\right) \\ $$$$\left({c}\right)\:\left(−\mathrm{2},−\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\left({d}\right)\left(−\infty,−\mathrm{2}\right)\mathrm{U}\left(\mathrm{2},\infty\right) \\ $$
Answered by einsteindrmaths@hotmail.fr last updated on 03/Apr/19
![{x}∈[0.1[ so is the same to say −3y^2 +2y+a^2 =0 has no solution] in[0.1[the solution of −3y^2 +2y+a^2 =0 are [−2+(√((4+12a^2 )))]/−6 and [−2−(√((4+12a^2 ) ))]/−6 only(2+(√((4+12a^2 ))) )/6 is greater than]zer so we want [2+(√((4+12a^2 ))) ]/6 <1=>(√((4+12a^2 )))<4===>12a^2 <12](Q57395.png)
$$\left\{{x}\right\}\in\left[\mathrm{0}.\mathrm{1}\left[\:{so}\:{is}\:{the}\:{same}\:{to}\:{say}\:\:−\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}{y}+{a}^{\mathrm{2}} =\mathrm{0}\:{has}\:{no}\:{solution}\right]\right. \\ $$$${in}\left[\mathrm{0}.\mathrm{1}\left[{the}\:{solution}\:{of}\:−\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}{y}+{a}^{\mathrm{2}} =\mathrm{0}\:{are}\:\left[−\mathrm{2}+\sqrt{\left(\mathrm{4}+\mathrm{12}{a}^{\mathrm{2}} \right)}\right]/−\mathrm{6}\:{and}\right.\right. \\ $$$$\left.\left[\left.−\mathrm{2}−\sqrt{\left(\mathrm{4}+\mathrm{12}{a}^{\mathrm{2}} \right)\:\:}\right]/−\mathrm{6}\:\:{only}\left(\mathrm{2}+\sqrt{\left(\mathrm{4}+\mathrm{12}{a}^{\mathrm{2}} \right.}\right)\:\:\right)/\mathrm{6}\:{is}\:{greater}\:{than}\right]{zer} \\ $$$${so}\:{we}\:{want}\:\:\:\left[\mathrm{2}+\sqrt{\left(\mathrm{4}+\mathrm{12}{a}^{\mathrm{2}} \right)}\:\right]/\mathrm{6}\:<\mathrm{1}=>\sqrt{\left(\mathrm{4}+\mathrm{12}{a}^{\mathrm{2}} \right)}<\mathrm{4}===>\mathrm{12}{a}^{\mathrm{2}} <\mathrm{12} \\ $$
Commented by rahul 19 last updated on 03/Apr/19
$${thanks}\:{sir}! \\ $$