If b∈Z^+ ∀ both the roots of equation x^2 −bx+132=0 are integers. Find (1)the largest possible value of b (2)the smallest possible value of b.


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Question Number 111035 by ZiYangLee last updated on 01/Sep/20

$If b \in Z^{+} \forall both the roots of equation$ $x^{2} - b x + 132 = 0 are integers .$ $Find$ $(1) the largest possible value of b$ $(2) the smallest possible value of b .$
	Answered by 1549442205PVT last updated on 02/Sep/20
	$x^2 −bx+132=(x−m)(x−n) ⇔ { ((m+n=b>0)),((mn=132=3.11.4)) :} WLOG suppose m≤n ⇒m∈{1,2,3,4,6,11} i)m=1⇒n=132⇒m+n=133 ii)m=2⇒n=66⇒m+n=68 iii)m=3⇒n=44⇒m+n=47 iv)m=4⇒n=33⇒m+n=37 v)m=6⇒n=22⇒m+n=28 vi)m=11⇒n=12⇒m+n=23 Thus,b=m+n get smallest value equal to 23 so that the given eq. x^2 −bx+132 has both its roots are integers$
	$x^{2} - bx + 132 = (x - m) (x - n)$ $\Leftrightarrow {\begin{cases} m + n = b > 0 \\ mn = 132 = 3.11.4 \end{cases}$ $WLOG suppose m ⩽ n$ $\Rightarrow m \in {1, 2, 3, 4, 6, 11}$ $i) m = 1 \Rightarrow n = 132 \Rightarrow m + n = 133$ $ii) m = 2 \Rightarrow n = 66 \Rightarrow m + n = 68$ $iii) m = 3 \Rightarrow n = 44 \Rightarrow m + n = 47$ $iv) m = 4 \Rightarrow n = 33 \Rightarrow m + n = 37$ $v) m = 6 \Rightarrow n = 22 \Rightarrow m + n = 28$ $vi) m = 11 \Rightarrow n = 12 \Rightarrow m + n = 23$ $Thus, b = m + n get smallest value$ $equal to 23 so that the given eq .$ $x^{2} - bx + 132 has both its roots are$ $integers$
	Commented by 1549442205PVT last updated on 02/Sep/20

	$Since m + n = b > 0, but mn = 132, so$ $both are positive, Sir!$
	Commented by Her_Majesty last updated on 01/Sep/20

	$m > 0 \land n > 0 o r m < 0 \land n < 0 \Rightarrow c h e c k y o u r$ $a n s w e r . t h e m e t h o d i s o k$
	Commented by Her_Majesty last updated on 02/Sep/20

	$s o r r y I m i s s e d t h a t b \in Z^{+}, y o u a r e r i g h t$
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