Question

Sa se afle ecuatiile dupa aducerea la forma cea mai simpla:
a) (2z+1) (z+3) - 5(z-2) -(z+1) (z-1)=0
b) (z+7) (2z+7)+ (z-1) (z) +7=0

Answer 1

a) $2 z^{2} +6z+z+3-5z+10-z ^{2} +1=0$
$z^{2} +2z+14=0$
Δ$= b^{2} -4ac= 2^{2} -4*14=4-56=-52$
$z_{1} = \frac{-b-i \sqrt{-delta} }{2a} = \frac{-4-i \sqrt{52} }{2} = \frac{-4-2i \sqrt{13} }{2} = \frac{2(-2-i \sqrt{13}) }{2} =-2-i \sqrt{13}$
$z_{2} = \frac{-b+i \sqrt{-delta} }{2a} = \frac{-4+i \sqrt{52} }{2} = \frac{-4+2i \sqrt{13} }{2} = \frac{2(-2+i \sqrt{13}) }{2} =-2+i \sqrt{13}$

S={$-2-i \sqrt{13}$;$-2+i \sqrt{13}$}

b)$2 z^{2} +7z+14z+49+ z^{2} -z+7=0$
$3 z^{2} +20z+56=0$
Δ$= b^{2} -4ac= 20^{2} -4*3*56=400-672=-272$
$z_{1} = \frac{-b-i \sqrt{-delta} }{2a} = \frac{-20-i \sqrt{272} }{6}= \frac{-20-4i \sqrt{17} }{6} = \frac{2(-10-2i \sqrt{17}) }{6} =$$\frac{-10-2i \sqrt{17} }{3}$
$z_{1} = \frac{-b+i \sqrt{-delta} }{2a} = \frac{-20+i \sqrt{272} }{6}= \frac{-20+4i \sqrt{17} }{6} = \frac{2(-10+2i \sqrt{17}) }{6} =$$\frac{-10+2i \sqrt{17} }{3}$

S={$\frac{-10-2i \sqrt{17} }{3}$;$\frac{-10+2i \sqrt{17} }{3}$}