If aεR and the equation : −3{x}^2 +2{x}+a^2 =0 has no integral solution, then all possible value of a lie in the interval : (a)(−1,0)U(0,1) (b)(1,2) (c) (−2,−1) (d)(−∞,−2)U(2,∞)


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Question Number 57389 by rahul 19 last updated on 03/Apr/19
$If aεR and the equation : −3{x}^2 +2{x}+a^2 =0 has no integral solution, then all possible value of a lie in the interval : (a)(−1,0)U(0,1) (b)(1,2) (c) (−2,−1) (d)(−∞,−2)U(2,∞)$
$I f a ϵ R a n d t h e e q u a t i o n :$ $- 3 {x}^{2} + 2 {x} + a^{2} = 0 h a s n o i n t e g r a l$ $s o l u t i o n, t h e n a l l p o s s i b l e v a l u e o f a$ $l i e i n t h e i n t e r v a l :$ $(a) (- 1, 0) U (0, 1) (b) (1, 2)$ $(c) (- 2, - 1) (d) (- \infty, - 2) U (2, \infty)$
	Answered by einsteindrmaths@hotmail.fr last updated on 03/Apr/19
	${x}∈[0.1[ so is the same to say −3y^2 +2y+a^2 =0 has no solution] in[0.1[the solution of −3y^2 +2y+a^2 =0 are [−2+(√((4+12a^2 )))]/−6 and [−2−(√((4+12a^2 ) ))]/−6 only(2+(√((4+12a^2 ))) )/6 is greater than]zer so we want [2+(√((4+12a^2 ))) ]/6 <1=>(√((4+12a^2 )))<4===>12a^2 <12$
	${x} \in [0.1 [s o i s t h e s a m e t o s a y - 3 y^{2} + 2 y + a^{2} = 0 h a s n o s o l u t i o n]$ $i n [0.1 [t h e s o l u t i o n o f - 3 y^{2} + 2 y + a^{2} = 0 a r e [- 2 + \sqrt{(4 + 12 a^{2})}] / - 6 a n d$ $[- 2 - \sqrt{(4 + 12 a^{2})}] / - 6 o n l y (2 + \sqrt{(4 + 12 a^{2}})) / 6 i s g r e a t e r t h a n] z e r$ $s o w e w a n t [2 + \sqrt{(4 + 12 a^{2})}] / 6 < 1 => \sqrt{(4 + 12 a^{2})} < 4 ===> 12 a^{2} < 12$
	Commented by rahul 19 last updated on 03/Apr/19

	$t h a n k s s i r!$
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