Question Number 144517 by Khalmohmmad last updated on 26/Jun/21
Answered by som(math1967) last updated on 26/Jun/21
![a^4 +b^4 +2a^2 b^2 −(a^2 +b^2 )−6=0 (a^2 +b^2 )^2 −(a^2 +b^2 )−6=0 x^2 −x−6=0 [let x=a^2 +b^2 ] (x−3)(x+2)=0 x=3 ,−2 ∴a^2 +b^2 =3,−2](https://www.tinkutara.com/question/Q144518.png)
$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{6}=\mathrm{0} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{6}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}=\mathrm{0}\:\:\:\left[{let}\:{x}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right] \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$${x}=\mathrm{3}\:,−\mathrm{2} \\ $$$$\therefore{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{3},−\mathrm{2} \\ $$