^2 - do kwadratu f(x)=x^2+x-30 Δ=121 x1=-6 x2=5 postać iloczynowa f(x)=(x+6)(x-5) f(x)=-5x^2+3x+2 Δ=49 x1=1 x2=-0,4 f(x)=-5(x+0,4)(x-1) zad 2 3(x+2)^2-2(x-2)(x+3)=8 3(x^2+4x+4)-2(x^2+3x-2x-6)=8 3x^2+12x+12-2x^2-2x+12=8 x^2+10x+16=0 Δ=36 x1=-8 x2=-2
1. p=-1/2 p=-b/2a q= -delta/4a delta= 1-4*1*(-30)=121 q=-121/4=-30,25 f(x)= (x+1/2)^2 -30,25 (postac kanoniczna) x1= (-1-11)/2=-6 x2=5 f(x)=(x-5)(x+6) - iloczynowa 2. p=-5/-10=0,5 delta= 3^2 - 4*(-5)*2=9+40=49 q=-49/-20=2,45 f(x)= -5(x-0,5)^2 + 2,45 - npostac kanoniczna x1=(-3-7)/-10=1 x2=0,4 f(x)=-5(x+0,4)(x-1) - iloczynowa 3. 3(x^2 +4x+4) -2(x^2 +3x -2x -6) -8 =0 3x^2 +12x +12 -2x^2 -6x +4x +12 -8=0 x^2 + 10x +16=0 delta= 100-64=36 pierw z delty = 6 x1=(-10-6)/2=-8 x2=(-10+6)/2=-2
