Solve-for-real-numbers-the-following-system-of-equations-a-a-1-b-1-a-2-b-3-2a-1-

Question Number 146854 by mathdanisur last updated on 16/Jul/21

S o l v e f o r r e a l n u m b e r s t h e f o l l o w i n g

s y s t e m o f e q u a t i o n s

{\begin{cases} a (a + 1) = b - 1 \\ a^{2} (b + 3) + 2 a = - 1 \end{cases}

Commented by Mrsof last updated on 16/Jul/21

a (a + b) + 4 = b + 3

a^{2} (a (a + 1) + 4) + 2 a = - 1

\Rightarrow a^{4} + a^{3} + 4 a^{2} + 2 a + 1 = 0

\Rightarrow a^{3} (a + 1) + 2 a^{2} + 2 a + 2 a^{2} + 1 = 0

\Rightarrow a^{3} (a + 1) + 2 a (a + 1) + (2 a^{2} + 1) = 0

\Rightarrow (a + 1) (a^{3} + 2 a) + (2 a^{2} + 1) = 0

h a s n o s o l u t i o n i n R

Commented by mathdanisur last updated on 16/Jul/21

t h a n k y o u S e r

Answered by liberty last updated on 16/Jul/21

{\begin{cases} a (a + 1) = b - 1 \Rightarrow b = a^{2} + a + 1 \\ a^{2} (b + 3) + 2 a = - 1 \end{cases}

\Rightarrow a^{2} (a^{2} + a + 4) + 2 a + 1 = 0

\Rightarrow a^{4} + a^{3} + 4 a^{2} + 2 a + 1 = 0

\Rightarrow (a^{2} + p a + 1) (a^{2} + q a + 1) =

a^{4} + a^{3} + 4 a^{2} + 2 a + 1

\Rightarrow a^{4} + (p + q) a^{3} + (2 + p q) a^{2} + (p + q) a + 1 =

a^{4} + a^{3} + 4 a^{2} + 2 a + 1

\Rightarrow {\begin{cases} p + q = 1 \\ 2 + p q = 4 \\ p + q = 2 \end{cases} t h e e q u a t i o n i n c o n s i s t e n t

Commented by mathdanisur last updated on 16/Jul/21

t h a n k y o u S e r