Question

sa se rezolve ecuatia z+$\frac{25}{z}$=6

Answer 1

${ }^{z)} z + \frac{25}{z} = {}^{z)} 6 \\ \\ \frac{ {z}^{2} }{z} + \frac{25}{z} = \frac{6z}{z} \\ \\ {z}^{2} + 25 = 6z \\ z {}^{2} - 6z + 25 = 0$

a=1

b=-6

c=25

delta=b²-4ac

delta=(-6)²-4•1•25=36-100=-64=64i²

z1= -b-✓delta/2a

z1= - (-6)-✓64i²/2•1= 6-8i/2 = 2(3-4i)/2=3-4i

z2=-b+✓delta/2a

z2= -(-6)+✓64i²/2•1= 6+8i/2= 3+4i

Answer 2

$\it z+\dfrac{25}{z}=6|_{\cdot z} \Rightarrow z^2+25=6z \Rightarrow z^2-6z+25=0 \Rightarrow z^2-6z+9+16=0 \Rightarrow \\ \\ \\ (z-3)^2=-16 \Rightarrow \sqrt{(z-3)^2}=\sqrt{-16} \Rightarrow |z-3|=4i \Rightarrow z-3=\pm4i\\ \\ \\ z_1=3-4i,\ \ \ z_2=3+4i$