N1
a) D=49+300>0 => 2 korna
b) D = 1-40<0 => 0 kornei
N2
a) Po teoreme Vietta:
Σ = 11 x1=14
=>
Π = -42 x2=-3
b) 2x^2+5x+2=0
D=25-16=9
x=(-5+3)/4 = -1/2
x=(-5-3)/4 = -2
c) x^4-13x^2+36=0
Σ=13 x1=9
=>
Π=36 x2=4
N3
S=1/2 * a * b, gde a, b -kateti
42= a(a-5)*0,5=0,5a^2 - 2,5a
a^2-5a-84=0
Po teoreme Vietta:
Σ=5 x1= 12 => a=12, b= 12-5=7
=>
Π=-84 x2=-7 - ne podhodit po smislu
Otvet: 12; 7
N4
-x^{2} -3x+10=0
x^{2}+3x-10=0
\left \{ {{Summa = -3} \atop {Proizvedenie = -10 }} \right. => \left \{ {{x1=-5} \atop {x=2}} \right" alt="\frac{-x(x+2)+8-(x-2)}{(4-x^{2})} = 0; ODZ:x\neq 2;-2
-x^{2} -3x+10=0
x^{2}+3x-10=0
\left \{ {{Summa = -3} \atop {Proizvedenie = -10 }} \right. => \left \{ {{x1=-5} \atop {x=2}} \right" align="absmiddle" class="latex-formula">