α and β,are 2 roots of eq: ax^2 +bx+c=0 with conditions: { ((α^2 =β+b)),((β^2 =α+a)) :} find: c in terms of: a and b.


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Question Number 55039 by behi83417@gmail.com last updated on 16/Feb/19
$α and β,are 2 roots of eq: ax^2 +bx+c=0 with conditions: { ((α^2 =β+b)),((β^2 =α+a)) :} find: c in terms of: a and b.$
$α a n d β, a r e 2 r o o t s o f e q :$ ${a x}^{2} + b x + c = 0 w i t h c o n d i t i o n s :$ ${\begin{cases} α^{2} = β + b \\ β^{2} = α + a \end{cases}$ $f i n d : c i n t e r m s o f : a a n d b .$
Commented by mr W last updated on 17/Feb/19
$i think the question has no solution, because there are too many conditions. α and β should be roots of ax^2 +bx+c=0, but they are already given by eqn. system { ((α^2 =β+b)),((β^2 =α+a)) :} this is not possible with given a and b as shown below. α+β=−(b/a) αβ=(c/a) on one side: α^2 +β^2 =α+β+a+b (α+β)^2 −2αβ=α+β+a+b (b^2 /a^2 )−((2c)/a)=−(b/a)+(a+b) b^2 −2ac=−ab+(a+b)a^2 ⇒c=(((b−a^2 )(a+b))/(2a)) ...(i) but on the other side: α^2 −β^2 =β−α+a−b (α−β)(α+β)=β−α+a−b (α−β)(1+α+β)=a−b (α−β)(1−(b/a))=a−b α−β=a α^2 +β^2 −2αβ=a^2 (b^2 /a^2 )−((2c)/a)=a^2 ((2c)/a)=(b^2 /a^2 )−a^2 =(((b+a^2 )(b−a^2 ))/a^2 ) ⇒c=(((b−a^2 )(a^2 +b))/(2a)) ...(ii) (i) and (ii) are contradiction.$
$i t h i n k t h e q u e s t i o n h a s n o s o l u t i o n,$ $b e c a u s e t h e r e a r e t o o m a n y c o n d i t i o n s .$ $α a n d β s h o u l d b e r o o t s o f {a x}^{2} + b x + c = 0,$ $b u t t h e y a r e a l r e a d y g i v e n b y e q n . s y s t e m$ ${\begin{cases} α^{2} = β + b \\ β^{2} = α + a \end{cases}$ $t h i s i s n o t p o s s i b l e w i t h g i v e n a a n d b a s s h o w n b e l o w .$ $α + β = - \frac{b}{a}$ $α β = \frac{c}{a}$ $o n o n e s i d e :$ $α^{2} + β^{2} = α + β + a + b$ ${(α + β)}^{2} - 2 α β = α + β + a + b$ $\frac{b^{2}}{a^{2}} - \frac{2 c}{a} = - \frac{b}{a} + (a + b)$ $b^{2} - 2 a c = - a b + (a + b) a^{2}$ $\Rightarrow c = \frac{(b - a^{2}) (a + b)}{2 a} . . . (i)$ $b u t o n t h e o t h e r s i d e :$ $α^{2} - β^{2} = β - α + a - b$ $(α - β) (α + β) = β - α + a - b$ $(α - β) (1 + α + β) = a - b$ $(α - β) (1 - \frac{b}{a}) = a - b$ $α - β = a$ $α^{2} + β^{2} - 2 α β = a^{2}$ $\frac{b^{2}}{a^{2}} - \frac{2 c}{a} = a^{2}$ $\frac{2 c}{a} = \frac{b^{2}}{a^{2}} - a^{2} = \frac{(b + a^{2}) (b - a^{2})}{a^{2}}$ $\Rightarrow c = \frac{(b - a^{2}) (a^{2} + b)}{2 a} . . . (i i)$ $(i) a n d (i i) a r e c o n t r a d i c t i o n .$
Commented by kaivan.ahmadi last updated on 17/Feb/19

$h i m r w$ $p l e a s e c h e c k t h e p r o o f a g a i n$ $α^{2} - β^{2} = (β - α) + (b - a) \Rightarrow$ $(α - β) (1 + (α + β)) = b - a$
Commented by mr W last updated on 17/Feb/19

$t h a n k y o u s i r! {y o u}^{'} r e r i g h t . i^{'} v e f i x e d .$ $p l e a s e c h e c k a g a i n .$
	Answered by kaivan.ahmadi last updated on 16/Feb/19
	${ ((α^2 −β=b)),((β^2 −α=a)) :}⇒(α^2 +β^2 )−(α+β)=a+b⇒ (α+β)^2 −2αβ−(α+β)=a+b⇒ (((−b)/a))^2 −((2c)/a)−(((−b)/a))=a+b⇒ ((2c)/a)=(b^2 /a^2 )+(b/a)−a−b⇒c=(b^2 /(2a))+(b/2)−(a^2 /2)−((ab)/2)= ((b^2 +ab−a^3 −a^2 b)/(2a))$
	${\begin{cases} α^{2} - β = b \\ β^{2} - α = a \end{cases} \Rightarrow (α^{2} + β^{2}) - (α + β) = a + b \Rightarrow$ ${(α + β)}^{2} - 2 α β - (α + β) = a + b \Rightarrow$ ${(\frac{- b}{a})}^{2} - \frac{2 c}{a} - (\frac{- b}{a}) = a + b \Rightarrow$ $\frac{2 c}{a} = \frac{b^{2}}{a^{2}} + \frac{b}{a} - a - b \Rightarrow c = \frac{b^{2}}{2 a} + \frac{b}{2} - \frac{a^{2}}{2} - \frac{a b}{2} =$ $\frac{b^{2} + a b - a^{3} - a^{2} b}{2 a}$
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