Given that a number is a factor of 144 and the square of the number added to five times the number is ≥ −6 find the number


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Question Number 35425 by Rio Mike last updated on 18/May/18

$G i v e n t h a t a n u m b e r i s a f a c t o r$ $o f 144 a n d t h e s q u a r e o f t h e n u m b e r$ $a d d e d t o f i v e t i m e s t h e n u m b e r$ $i s ⩾ - 6 f i n d t h e n u m b e r$
	Answered by Rasheed.Sindhi last updated on 19/May/18
	$n∣144 ∧ n^2 +5n≥−6 n^2 +5n+6≥0 (n+2)(n+3)≥0 (n+2)(n+3)=0 ∣ ^∗ (n+2)(n+3)>0 n=−2 ∣ n=−3.........I ^∗ (n+2)(n+3)>0 n+2>0 ∧ n+3>0 ∣ n+2<0 ∧ n+3<0 n>−2 ∧ n>−3 ∣ n<−2 ∧ n<−3 n>−2 ∣ n<−3...........II I & II: n≥−2 or n≤−3 Hence Factors of 144 ≥−2 or ≤−3 {,..−8−6 ,−4,−3,−2,−1,1,2,3,4,6,8....} ={±1,±2,±3,±4,±6,±8,±9,....} Or All the factors of 144$
	$n ∣ 144 \land n^{2} + 5 n ⩾ - 6$ $n^{2} + 5 n + 6 ⩾ 0$ $(n + 2) (n + 3) ⩾ 0$ $(n + 2) (n + 3) = 0 ∣^{} (n + 2) (n + 3) > 0$ $n = - 2 ∣ n = - 3 . . . . . . . . . I$ $^{} (n + 2) (n + 3) > 0$ $n + 2 > 0 \land n + 3 > 0 ∣ n + 2 < 0 \land n + 3 < 0$ $n > - 2 \land n > - 3 ∣ n < - 2 \land n < - 3$ $n > - 2 ∣ n < - 3 . . . . . . . . . . . II$ $I & II : n ⩾ - 2 or n ⩽ - 3$ $Hence$ $Factors of 144 ⩾ - 2 or ⩽ - 3$ ${, . . - 8 - 6, - 4, - 3, - 2, - 1, 1, 2, 3, 4, 6, 8 . . . .}$ $= {\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 9, . . . .}$ $Or All the factors of 144$
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