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Question Number 62672 by bshahid010@gmail.com last updated on 24/Jun/19

Commented by Rasheed.Sindhi last updated on 24/Jun/19

$b \in I, I m e a n s i r r a t i o n a l ?$
Commented by mr W last updated on 24/Jun/19

$i t h i n k h e m e a n s i n t e g e r s .$
Commented by MJS last updated on 24/Jun/19

$I {don}^{'} t understand . what does ‘ ‘ {independent}^{″}$ $mean ? what does b \in I mean ?$
Commented by mr W last updated on 24/Jun/19

$r a s h e e d s i r : c a n y o u g i v e a s o l u t i o n$ $t o t h i s q u e s t i o n ? a s s u m e b = i n t e g e r .$
Commented by bshahid010@gmail.com last updated on 24/Jun/19

$Here I means integer which . . . I denotes the set of Integers .$ $Independent means that should not be related to each$ $other anyways$
Commented by Rasheed.Sindhi last updated on 24/Jun/19

$C o u n t e r E x a m p l e$ $a = 8 (i n t e g e r), b = 3 (o d d p r i m e), c = 5 (o d d p r i m e$ $b^{2} - 4 a c = 8^{2} - 4 (3) (5) = 4 p e r f e c t s q u a r e$ $\Rightarrow R o o t s a r e r a t i o n a l$ $(Y e t b o t h r o o t s a r e d e p e n d a n t o n a, b, c$ $d e p e n d a n t i n a s e n s e t h a t t h e i r s o l u t i o n$ $i n v o l v e t h e a b o v e v a l u e s o f a, b & c ? ? ?)$ $x = \frac{- 3 \pm 2}{2 (8)} = - \frac{1}{16}, - \frac{5}{16}$ $A c t u a l l y r o o t s a r e a l w a y s d e p e n d a n t$ $o n a, b & c . S o {i t}^{'} s q u e s t i o n a b l e t o s a y$ $a b o u t r o o t s w h i c h a r e i n d e p e n d a n t$ $o f a, b, c .$
Commented by mr W last updated on 24/Jun/19

$i t h i n k i t i s m e a n t t h a t o n e o f t h e$ $r o o t s i s a c o n s t a n t n o m a t t e r w h i c h$ $v a l u e s a, b, c a r e o f .$ $i n y o u r e x a m p l e :$ $a = 3, b = 8, c = 5$ $b^{2} - 4 a c = 4$ $x = \frac{- 8 \pm 2}{6} = - 1, - \frac{5}{3}$ $i . e . x = - 1, - \frac{c}{a}$ $i t h i n k o n e r o o t i s a l w a y s - 1,$ $i n d e p e n d e n t l y f r o m t h e v a l u e s o f$ $a, b, c .$
Commented by Rasheed.Sindhi last updated on 24/Jun/19

$S i r, t h e s e t w o e x a m p l e s p r o v e t h a t$ $n e i t h e r o f t h e r o o t s i s c o n s t a n t, a l t h o u g h$ $w e h a v e f u l f i l l e d a l l t h e r e q u i r e m e n t s :$ $b i n t e g e r, a, c o d d p r i m e s a n d r a t i o n a l$ $r o o t s .$ $b = 8, a = 3, b = 5$ $b = 8, a = 5, b = 3$ $\Rightarrow I n b o t h c a s e s r o o t s a r e r a t i o n a l$ $B u t n o r o o t i s s a m e!$
Commented by MJS last updated on 24/Jun/19

$a = p c = p \pm 2 n$ $b = \pm 2 (n + p) \Rightarrow x = \mp 1 \lor x = \mp \frac{2 n + p}{p}$ $b = \pm [p (2 n + p) + 1] \Rightarrow x = \mp (2 n + p) \lor x = \mp \frac{1}{p}$
	Answered by mr W last updated on 24/Jun/19
	$x=((−b±(√(b^2 −4ac)))/(2a))=((−b±n)/(2a)) b^2 −4ac=n^2 , say b^2 −n^2 =4ac (b−n)(b+n)=4ac=2^2 a^1 c^1 ⇒b and n must be both even or odd. i.e. b−n and b+n must be even. since a and c are odd primes, RHS has 3×2×2=12 factors, or it can be expressed as the product of two numbers in 6 ways: 4ac=1×4ac=2×2ac=4×ac=a×4c=2a×2c=4a×c case 1: { ((b−n=1 ⇒ bad!)),((b+n=4ac)) :} case 2: { ((b−n=2 ⇒ b=ac+1, n=ac−1)),((b+n=2ac)) :} case 2′: { ((b−n=2ac ⇒ b=ac+1, n=1−ac)),((b+n=2)) :} case 3: { ((b−n=4)),((b+n=ac ⇒ bad!)) :} case 4: { ((b−n=a ⇒ bad!)),((b+n=4c)) :} case 5: { ((b−n=2a ⇒ b=c+a, n=c−a)),((b+n=2c)) :} case 5′: { ((b−n=2c ⇒ b=c+a, n=a−c)),((b+n=2a)) :} case 6: { ((b−n=4a)),((b+n=c ⇒ bad!)) :} we see only case 2 and case 5 are suitable. from case 2+2′: b=ac+1 x=((−b±n)/(2a))=((−ac−1±(1−ac))/(2a))= { ((−(1/a))),((−c)) :} from case 5+5′: b=a+c x=((−b±n)/(2a))=((−c−a±(c−a))/(2a))= { ((−1)),((−(c/a))) :} summary: to fulfill the condictions, b must be ac+1 or a+c. with b=a+c, one root is always −1. but with b=ac+1, both roots are dependent from a and c, that is to say the question is not correct!$
	$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- b \pm n}{2 a}$ $b^{2} - 4 a c = n^{2}, s a y$ $b^{2} - n^{2} = 4 a c$ $(b - n) (b + n) = 4 a c = 2^{2} a^{1} c^{1}$ $\Rightarrow b a n d n m u s t b e b o t h e v e n o r o d d .$ $i . e . b - n a n d b + n m u s t b e e v e n .$ $s i n c e a a n d c a r e o d d p r i m e s, R H S$ $h a s 3 \times 2 \times 2 = 12 f a c t o r s, o r i t c a n$ $b e e x p r e s s e d a s t h e p r o d u c t o f t w o$ $n u m b e r s i n 6 w a y s :$ $4 a c = 1 \times 4 a c = 2 \times 2 a c = 4 \times a c = a \times 4 c = 2 a \times 2 c = 4 a \times c$ $c a s e 1 : {\begin{cases} b - n = 1 \Rightarrow b a d! \\ b + n = 4 a c \end{cases}$ $c a s e 2 : {\begin{cases} b - n = 2 \Rightarrow b = a c + 1, n = a c - 1 \\ b + n = 2 a c \end{cases}$ $c a s e 2^{'} : {\begin{cases} b - n = 2 a c \Rightarrow b = a c + 1, n = 1 - a c \\ b + n = 2 \end{cases}$ $c a s e 3 : {\begin{cases} b - n = 4 \\ b + n = a c \Rightarrow b a d! \end{cases}$ $c a s e 4 : {\begin{cases} b - n = a \Rightarrow b a d! \\ b + n = 4 c \end{cases}$ $c a s e 5 : {\begin{cases} b - n = 2 a \Rightarrow b = c + a, n = c - a \\ b + n = 2 c \end{cases}$ $c a s e 5^{'} : {\begin{cases} b - n = 2 c \Rightarrow b = c + a, n = a - c \\ b + n = 2 a \end{cases}$ $c a s e 6 : {\begin{cases} b - n = 4 a \\ b + n = c \Rightarrow b a d! \end{cases}$ $w e s e e o n l y c a s e 2 a n d c a s e 5 a r e s u i t a b l e .$ $f r o m c a s e 2 + 2^{'} :$ $b = a c + 1$ $x = \frac{- b \pm n}{2 a} = \frac{- a c - 1 \pm (1 - a c)}{2 a} = {\begin{cases} - \frac{1}{a} \\ - c \end{cases}$ $f r o m c a s e 5 + 5^{'} :$ $b = a + c$ $x = \frac{- b \pm n}{2 a} = \frac{- c - a \pm (c - a)}{2 a} = {\begin{cases} - 1 \\ - \frac{c}{a} \end{cases}$ $s u m m a r y :$ $t o f u l f i l l t h e c o n d i c t i o n s, b m u s t$ $b e a c + 1 o r a + c . w i t h b = a + c, o n e$ $r o o t i s a l w a y s - 1 . b u t w i t h b = a c + 1,$ $b o t h r o o t s a r e d e p e n d e n t f r o m a a n d c,$ $t h a t i s t o s a y t h e q u e s t i o n i s n o t$ $c o r r e c t!$
	Commented by Rasheed.Sindhi last updated on 24/Jun/19

	$M a t c h l e s s a p p r o a c h S i r!$
	Commented by mr W last updated on 24/Jun/19

	$t h a n k s s i r! b u t i j u s t p r o v e d t h a t t h e$ $q u e s t i o n i s n o t c o r r e c t .$
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