Question

Va rog ecuația asta ​

Answer 1

Răspuns:

$x^2-(4-i)x+5-5i=0$

$x1,x2=\frac{4-i}{2}$±$\frac{\sqrt{(4-i)^2-4(5-5i)} }{2}$=$\frac{4-i}{2}$±$\frac{\sqrt{-5+12i} }{2}$

Fie $z^2=-5+12i=>z=\sqrt{-5+12i}$, unde z este un numar complex de forma $z=a+bi$

$z^2=a^2-b^2+2abi=-5+12i$

$=>a^2-b^2=-5$

$=>2ab=12=>a=\frac{6}{b}$ (Inlocuim in prima ecuatie)

$(\frac{6}{b} )^2-b^2=-5<=>\frac{36}{b^2}-b^2+5=0|*b^2$

$36-b^4+5b^2=0;$Fie $t=b^2$$=>-t^2+5t+36=0=>t1,t2=\frac{-5}{-2}$±$\frac{\sqrt{5^2-4*36*(-1)} }{-2}$

$=> t1=-4;t2=9$

Ne intoarcem la afirmatia facuta mai devreme $t=b^2$

$-4=b^2=>b$∉$R$, iar coeficientul imaginar nu poate fi complex, asta inseamna ca nu avem solutii pentru b cand t =-4

$9=b^2=>b=$ ±$3$, din nou, ne intoarcem la afirmatia facuta $a=\frac{6}{b} =>a1=-2,a2=2$

$=>z1=2+3i;z2=-2-3i$

Acum ne intoarcem la Δ in ecuatia de gr. 2

$x1,x2=\frac{4-i}{2}$ ±$\frac{2+3i}{2}$$=>x1=\frac{4-i+2+3i}{2}=\frac{6+2i}{2}=3+i$

$x2=\frac{4-i-2-3i}{2}=\frac{2-4i}{2}=1-2i$