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

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Question Number 146854 by mathdanisur last updated on 16/Jul/21
Solve for real numbers the following system of equations { ((a(a+1) = b−1)),((a^2 (b+3)+2a = −1)) :}
$${Solve}\:{for}\:{real}\:{numbers}\:{the}\:{following} \\ $$$${system}\:{of}\:{equations} \\ $$$$\begin{cases}{{a}\left({a}+\mathrm{1}\right)\:=\:{b}−\mathrm{1}}\\{{a}^{\mathrm{2}} \left({b}+\mathrm{3}\right)+\mathrm{2}{a}\:=\:−\mathrm{1}}\end{cases} \\ $$
Commented by Mrsof last updated on 16/Jul/21
a(a+b)+4=b+3 a^2 (a(a+1)+4)+2a=−1 ⇒a^4 +a^3 +4a^2 +2a+1=0 ⇒a^3 (a+1)+2a^2 +2a+2a^2 +1=0 ⇒a^3 (a+1)+2a(a+1)+(2a^2 +1)=0 ⇒(a+1)(a^3 +2a)+(2a^2 +1)=0 has no solution in R
$${a}\left({a}+{b}\right)+\mathrm{4}={b}+\mathrm{3} \\ $$$$ \\ $$$${a}^{\mathrm{2}} \left({a}\left({a}+\mathrm{1}\right)+\mathrm{4}\right)+\mathrm{2}{a}=−\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{a}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{a}^{\mathrm{3}} \left({a}+\mathrm{1}\right)+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{a}^{\mathrm{3}} \left({a}+\mathrm{1}\right)+\mathrm{2}{a}\left({a}+\mathrm{1}\right)+\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\left({a}+\mathrm{1}\right)\left({a}^{\mathrm{3}} +\mathrm{2}{a}\right)+\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$ \\ $$$${has}\:{no}\:{solution}\:{in}\:{R} \\ $$
Commented by mathdanisur last updated on 16/Jul/21
thank you Ser
$${thank}\:{you}\:{Ser} \\ $$
Answered by liberty last updated on 16/Jul/21
{ ((a(a+1)=b−1⇒b=a^2 +a+1)),((a^2 (b+3)+2a=−1)) :} ⇒a^2 (a^2 +a+4)+2a+1=0 ⇒a^4 +a^3 +4a^2 +2a+1=0 ⇒(a^2 +pa+1)(a^2 +qa+1)= a^4 +a^3 +4a^2 +2a+1 ⇒a^4 +(p+q)a^3 +(2+pq)a^2 +(p+q)a+1= a^4 +a^3 +4a^2 +2a+1 ⇒ { ((p+q=1)),((2+pq=4)),((p+q=2)) :} the equation inconsistent
$$\:\begin{cases}{{a}\left({a}+\mathrm{1}\right)={b}−\mathrm{1}\Rightarrow{b}={a}^{\mathrm{2}} +{a}+\mathrm{1}}\\{{a}^{\mathrm{2}} \left({b}+\mathrm{3}\right)+\mathrm{2}{a}=−\mathrm{1}}\end{cases} \\ $$$$\:\Rightarrow{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{a}+\mathrm{4}\right)+\mathrm{2}{a}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{a}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{pa}+\mathrm{1}\right)\left({a}^{\mathrm{2}} +{qa}+\mathrm{1}\right)= \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{4}} +{a}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +\left({p}+{q}\right){a}^{\mathrm{3}} +\left(\mathrm{2}+{pq}\right){a}^{\mathrm{2}} +\left({p}+{q}\right){a}+\mathrm{1}= \\ $$$$\:\:\:\:{a}^{\mathrm{4}} +{a}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1} \\ $$$$\Rightarrow\begin{cases}{{p}+{q}=\mathrm{1}}\\{\mathrm{2}+{pq}=\mathrm{4}}\\{{p}+{q}=\mathrm{2}}\end{cases}\:{the}\:{equation}\:{inconsistent} \\ $$
Commented by mathdanisur last updated on 16/Jul/21
thank you Ser
$${thank}\:{you}\:{Ser} \\ $$

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