and-are-2-roots-of-eq-ax-2-bx-c-0-with-conditions-2-b-2-a-find-c-in-terms-of-a-and-b-

Question Number 55039 by behi83417@gmail.com last updated on 16/Feb/19

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 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} \dots (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} \dots (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

{\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}