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Question Number 151490 by mathdanisur last updated on 21/Aug/21

 { ((a^2 =2a+b)),((b^2 =a+2b)) :}  and  a≠b  find  (√(a^2 +b^2 +1))

$$\begin{cases}{\mathrm{a}^{\mathrm{2}} =\mathrm{2a}+\mathrm{b}}\\{\mathrm{b}^{\mathrm{2}} =\mathrm{a}+\mathrm{2b}}\end{cases}\:\:\mathrm{and}\:\:\mathrm{a}\neq\mathrm{b}\:\:\mathrm{find}\:\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{1}} \\ $$

Answered by iloveisrael last updated on 21/Aug/21

⇒a^2 +b^2 =3(a+b)  ⇒a^2 −b^2 =a−b ; (a+b)(a−b)−(a−b)=0  ⇒a+b=1 ⇒a^2 +b^2 =3  ⇒(√(a^2 +b^2 +1)) =2

$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{3}\left({a}+{b}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={a}−{b}\:;\:\left({a}+{b}\right)\left({a}−{b}\right)−\left({a}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}+{b}=\mathrm{1}\:\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{1}}\:=\mathrm{2} \\ $$

Commented by mathdanisur last updated on 21/Aug/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

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